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1! -.Vftbbcr- introo Southern Branch of the University of California Los Angeles QA This book is DUE on the '' l ^ ' stamped below r I, INTRODUCTORY MATHEMATICAL ANALYSIS BY W. PAUL WEBBER, PH.D. Assistant Professor of Mathematics in the University of Pittsburgh AND LOUIS CLARK PLANT, M.Sc. Professor of Mathematics in Michigan Agricultural College FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED 1919 COPYBIGHT, 1919, BY W. PAUL WEBBER AND LOUIS C. PLANT Stanbope jpress . H.GILSON COMPANY BOSTON, U.S.A. Asy Wsg PREFACE The present course is the result of several years of study and trial in the classroom in an effort to make an introduction to college mathematics more effective, rational and better suited to its place in a scheme of education under modern conditions of life. A broader field has been attempted than is customary in books of its class. This is made possible by certain principles which controlled the construction of the text. One principle on which the course is built is correlation by topics. For example, all methods of calculation have been associated in one chapter and early in the course in order to be- available for use in the sequel. The function idea has also been emphasized and used as a means of correlation. Brevity and directness of treatment have contributed to reduce the size of the book. An effort has been made to keep in view of the student the steps in the development of the subject and to point out useful contacts of mathematics with affairs. The first two chapters are intended to be used for review and reference at the discretion of the instructor. Graphic representation and its uses have been given consider- able attention. The simple cases of determining empirical formulae give a very valuable drill in the solution of simul- taneous equations and a foundation for later work in the laboratory. The treatment of the trigonometric functions is brief, direct and in some respects more advanced in style than is customary in current texts in trigonometry which are constructed mostly from the secondary school standpoint. Large use of the func- iv PREFACE tions is made in a variety of applications in immediately following chapters. More than usual attention is given to vectors. The value and convenience of vector methods in science and engineering seem to justify this emphasis. The part dealing with vector products and the problems depending on it may, however, be omitted without inconvenience in later chapters. The chapter on series may seem a little heavy for freshmen but it comes in the second half of the course and is directly applied to functions within the experience of the student in the preceding text. What is given on differential and integral calculus is intended as an introduction for those who are to take the regular and fuller course in calculus. For those who are not to continue their mathematics it will furnish an introduction to the methods of calculus and some important definite applications. The integral has first been regarded as the inverse of the derivative and nothing is said about the differential. This seems natural and in accord with the idea of the solution of differential equa- tions under many actual conditions where a function is sought whose derivative is given. Following, the integral is regarded as a summation of elements and some further applications are introduced. In the list of integrals for reference both the in- verse and the differential forms are given. In general no effort at rigor beyond reasonable conviction has been attempted. Proofs have been given for some theorems that many teachers may prefer to regard as assumptions. These proofs may, therefore, be omitted at the discretion of the teacher. A number of what appear as theorems in some texts are here given as exercises. For this reason it is recom- mended that each student be held for practically all the exter- cises appearing regularly through the text. Selections may be made at the instructor's discretion from the exercises at the end of each chapter. The teacher will find an opportunity for originality in develop- ing the text and at times a necessity for more details. PREFACE V The entire course has been given several times to classes meeting daily during a period equivalent to four terms of three months each. The work can be covered in less time. The authors take this opportunity to acknowledge their obligation to Dean H. B. Meller, of the University of Pittsburgh, for affording the opportunity to have the course tried out in the classroom and for a number of problems. If this book shall contribute toward making more satisfactory and more economical the modicum of college mathematics the authors will feel well repaid for the considerable labor of its composition. W. P. W. L. C. P. TABLE OF CONTENTS (Numbers refer to sections) CHAPTER I REVIEW OF ALGEBRA PAGE 1. Signs of aggregation 1 2. Index laws 2 3. Important cases of factoring 3 4. Lowest common multiple 4 5. Highest common factor 5 6. Fractions 7 7. Equations; root of an equation 8 8. Simultaneous equations and elimination 8 9. Formulas relating to radicals 11 10. Quadratic equations 12 11. Inequalities 14 12. Binomial formula 15 * CHAPTER II GEOMETRICAL THEOREMS 13. Geometrical theorems and formulas 16 CHAPTER III METHODS OF CALCULATION 14. Need of numbers and calculations 20 15. Important discoveries relating to numbers 15 16. Mechanical devices for calculating 21 17. Graphic or geometric representation of numbers; scale 22 18. Arithmetical operations by geometric methods 22 19. Idea of logarithms 24 20. Definition of logarithm 25 21. Rules for calculating with logarithms ' 25 22. Characteristic and mantissa of a logarithm 26 viii TABLE OF CONTENTS PAOE 23. Use of tables of logarithms in calculating 28 24. Exponential equations solved by use of logarithms 28 25. Logarithmic scale; slide rule 30 26. Rules for calculating with Mannheim slide rule 31 26a. Double interpolation 31 CHAPTER IV GRAPHIC REPRESENTATION 27. Graphic representation of statistical data 36 28. Axes and coordinates defined 37 CHAPTER V RATIO PROPORTION AND VARIATION 29. Proportion 45 30. Ratio; measurement 45 31. Formulas of proportion 46 32. Variation; direct, inverse and joint 46 CHAPTER VI RECTANGULAR COORDINATE SYSTEM; GRAPHIC REPRESEN- TATION OF EQUATIONS 33. Axes and coordinates; quadrants 52 34. Graphs of functions and equations 53 35. Emperical equations or formulas 57 CHAPTER VII NUMBERS, VARIABLES, FUNCTIONS AND LIMITS 36. Classes of numbers 62 37. Variable; function; sequence; limit 63 38. Functional notation 65 39. Problem of mathematics 66 40. Increment of a variable 66 41. Special forms and limits 66 42. Proofs of theorems 67 43. Limiting values of expressions for certain values of the variable 68 44. Idea of function developed from number pairs and curves 69 TABLE OF CONTENTS ix CHAPTER VIII CIRCULAR (TRIGONOMETRIC) FUNCTIONS AND THEIR APPLICATIONS PAGE 45. Problem 71 46. Angle defined 72 47. Trigonometric ratios defined 73 48. Fundamental formulas 75 49. Functions at the quadrant limits 78 50. Functions of negative angles 79 51. Complement relations 80 52. Reduction to functions of angles less than 90 81 53. Addition theorems 85 54. Trigonometric equations; inverse functions 89 55. Formulas relating to right triangles 91 56. Directions for solving problems 91 57. Sine law 95 58. Cosine law 95 59. Example of the use of the sine law 96 60. Example of the use of the cosine law 97 61. Conversion formulas 98 62. Tangent law 99 63. Ambiguous cases 100 64. Double and half angles 101 65. Angle of a triangle in terms of the sides 102 66. Radius of inscribed circle 103 67. Radius of circumscribed circle 103 68. Radian measure of angles 104 68a. Mil as a unit of angular measure 106 69. Explanatory definitions relating to field work 106 70. Graphs of trigonometric functions 108 71. Graphs of inverse functions 110 72. Equations involving trigonometric functions as unknowns 112 72a. Special interpolations and tables 113 CHAPTER IX POLAR COORDINATES; COMPLEX NUMBERS; VECTORS 73. Polar coordinates 120 74. Powers of V 1 = i 122 75. Geometrical representation of complex numbers 122 X TABLE OF CONTENTS PAGE 76. Arithmetical operations with complex numbers 123 77. Multiplication and division in polar form 124 78. Vector quantities; vectors 125 79. Rectangular and polar notations of vectors 125 80. Addition and subtraction of vectors 126 81. Multiplication of vectors 128 82. Components of vectors on axes 130 83. Equilibrium of particles and rigid bodies 132 CHAPTER X EQUATIONS 84. Integral equation; factor theorem 139 85. Fundamental theorem of algebra 139 86. Identity theorem 140 87. Remainder theorem 141 88. Zero and infinite roots 141 89. Synthetic division 142 90. Theorem on root of an equation 143 91. Solution of numerical equations 144 92. Quadratic equations 149 93. Equations in quadratic form 150 CHAPTER XI LINEAR FUNCTIONS AND THE STRAIGHT LINE 94. General linear function 152 95. Theorem on the linear equation and the straight line 152 96. Families of straight lines 153 97. Converse theorem 154 98. Normal form of the equation of the straight line 156 99. Different forms of equation of the straight line 157 100. Distance between two points 158 101. Division of a line segment 158 102. Angle between two lines 160 CHAPTER XII EQUATIONS OF THE SECOND AND HIGHER DEGREES AND THEIR GRAPHS 103. Explicit and implicit functions defined 1 63 104. Explicit quadratic function; discussion of a curve 163 TABLE OF CONTENTS xi PAGE 105. Implicit quadratic function 165 106. Reference to conic sections 166 107. Functions of the third degree 167 108. Rational fractional function 167 109. Irrational function 168 110. Simultaneous equations of the second and higher degrees 169 111. Equivalence of equations 169 112. Some systems of quadratic equations 171 CHAPTER XIII TRANSFORMATION OF COORDINATES 113. Linear transformation; rotating the axes; moving the origin. . . 174 CHAPTER XIV CONIC SECTIONS 114. Conic sections defined 177 115. Parabola 177 116. Ellipse 179 117. Hyperbola 181 118. Diameters 183 119. Eccentric angle 185 120. General equation of the second degree in two variables 187 121. Confocal conies 188 122. Centers of conies 189 CHAPTER XV THEOREMS ON LIMITS; DERIVATIVES AND THEIR APPLICATIONS 123. Limit of sum of infinitesimals 193 124. Limit of sum of variables 193 125. Limit of product of variables ,. 194 126. Limit of quotient of two variables 194 127. Definition and formation of derivative 194 128. Reference to special rules 197 129. Derivative of constant 197 130. Derivative of variable with respect to itself 197 131. Derivative of positive integral power of function 198 132. Extension to positive fractional power 198 xii TABLE OF CONTENTS PAGE 133. Extension to negative power 198 134. Extension to irrational power 199 135. Extension to imaginary power 199 136. Derivative of product of functions 199 137. Derivative of quotient 200 138. Derivative from an implicit function 200 139. Slope of a curve at a point; equation of tangent line 202 140. Maximum and minimum values of functions 205 141. Use of derivative to determine maximum and minimum values 205 142. Increasing and decreasing functions; conditions for maximum and minimum values 207 143. Use of second derivative to determine maximum and minimum values .' 208 144. Use of derivative to define motion 214 145. Use of derivative to discover equal roots of equations 217 146. Force the derivative of momentum with respect to time 218 CHAPTER XVI SERIES; TRANSCENDENTAL FUNCTIONS 147. Sequences 219 148. Series defined; convergence defined 219 149. Arithmetic series 220 149a. Geometric series 222 150. Special case of geometric series 224 151 . Harmonic series 225 152. Convergence of series . . '. 226 153. Comparison test of convergence 228 154. Standard series for comparison 228 155. Ratio test of convergence 230 156. Series with complex terms 231 157. Expansion of functions in power series 232 158. Functions expanded about the origin; functions expanded about any point; formulas 232 159. Binomial expansion, any exponent. 234 160. Exponential function and series 236 161. Theorem on logarithms 236 162. Derivative of exponential function; derivative of logarithmic function 237 163. Use of logarithm in calculating derivatives 238 164. Logarithmic series; calculation of logarithms 239 165. Exponential (Euler's) values of sine and cosine 240 TABLE OF CONTENTS xiii PAGE 166. Derivatives of sine and cosine 241 167. Expansion of sine and cosine in series 242 167a. Logarithmic graphs 244 167b. Empirical formulas derived from logarithmic graph 246 167c. Empirical formulas derived from semilogarithmic graph 247 CHAPTER XVII INTEGRATION 168. Integral defined 250 169. Directions for solving problems 253 170. Definite integral 254 171. Area under a curve 256 172. Volume of solid of revolution 257 173. Average value of a function over an interval of the variable. . . 259 174. Work of a variable force 261 175. Integral regarded as a sum of infinitesimal elements 263 176. Centroids determined by integration 264 177. Integration by parts 267 178. Length of an arc of a curve 268 179. Simplifying integrand by transformation of variable 269 Additional formulas 271 Supplementary exercises 271 Table of integrals 276 Use of tables 280 Calculating tables 284 INTRODUCTORY MATHEMATICAL ANALYSIS CHAPTER I REVIEW OF ALGEBRA 1. Signs of aggregation are used to indicate that an opera- tion is to extend to each term of a group of terms or to separate or to specify torms or factors in an algebraic expression. Thus: 1. 3 a (6 + c + d) shows that each term inside the " ( ) " is to be multiplied by the factor, 3 a, which precedes the " ( ) ", or 3 a (b + c + d) is equivalent to 3 ab + 3 ac + 3 ad. 2. V6 a + 3 b - 24 shows by the " " (vinculum) that the "V" (square root) is to be taken of the expression under the " " considered as a whole, and not the square root of each term separately. 3. a (2 c 4 & + d) or a 2 c 4 6 + d shows that the terms in '' ( ) " or under " " are all to be subtracted from a. Since we may either subtract these singly or first combine them and subtract the combined result, we may express the operation as, = a-2c- (-46) -d=a-2c + 4&-d. 4. (3 -f- 6 a 4 6) (a 5 c + d) shows that the combined value of the terms in each " ( ) " is to be multiplied by the other. This is accomplished by multiplying every term in one " ( ) " by each term in the other " ( ) " and reducing the result to the simplest form. 5. a -\- b c + d + Vc e may, by using a sign of aggre- gation, say a " ( )," be written as a + & + d (c + e 1 2 REVIEW OF ALGEBRA Remark. When a sign of aggregation is preceded by a sign, the sign of aggregation may be removed mill us the si ns of the terms within the si ^ of aggregation. We must not confuse signs of aggregation with signs of operation. Thus, in the expression Va 6, the "V" is not a sign of aggregation, but shows that the square root is to be taken of the entire expression under the " considered as a whole. The minus sign before the "V" is not to be regarded as preceding the sign of aggregation and there- fore the " - " in the expression cannot be removed by the usual rule. The operation of square root must first be per- formed. Similarly, other cases are to be handled. I2x - (9x + 7x) = ?; ofc + 3 - a& = ? 2. xy - Vx z -2x+l = ? ; o% 2 - x 2 y 2 - 1 = ? 3. a (a + & (c d-\- e a) + c) 6 c + d e = ! 4. (ab ~(cd+ 1)) (06 - (cd - 1)) = ? 5. (2z 2 +(3z-l) (4 3 + 5)) (5z 2 -(4s + 3) (x - 2)) =? 6. x {-l2y-[2x+(-4:y- (-7x-5y) -Qx-9y 7. (a 2 - (6 2 + c 2 )) (a 2 - (6 2 - c 2 )) = ? 8. 20 a (10 a - (6 a - c - '(5 a ->)).+ c) = ? 9. (V9-6z 2 + z 4 - V25-10a + a 2 ) (3 - \/25) = 2. The index laws may be stated in the form of equations as follows: 1. a m d n = a m+n . 2. a m -i- a n = a m ~ n . 3. a m -i- a m = 1, also a m -5- a m = a w-m = a. ' .'. a = 1 for any value of a. 4. 1 -T- a m = a -f- a m = a~ m = o-, but 1 -i- a m = I/a*. /. a"* = l/a m . 5. (o m ) n = a mn . 60 (06) n = a n 6 n . CASES OF FACTORING 3 Perform the indicated operations with the expressions just as they are written and change the result so that all exponents shall be positive. 1 . (a 2 ar ! + 3 a 3 x~ 2 ) (4 a~ l - 5 or 1 + 6 cur 2 ) = ? 2. (a - a- 1 ) (a - cr 2 ) (a - a- 3 ) = ? 3. (x* - ?/*) (** + x V + r) = ? 4. (18 y~* + 23 + xr*y + 6arV) 4- (3 xV'+z* - 2 x~*y) = ? 5. (x-27/) 6 = ? 6. (3 alb* + 4 a&3 - a ? 6) (6 o*H - 8 a~*H - 2 a~*) = ? 7. (8x 2 -2x-3)/(12x 2 -25x-12) =? When x = f , (a) by direct substitution, (6) by carrying out the division and substituting in the quotient. 8. (XP - 3 x?- 1 + 4 x*- 2 - 6 x"- 3 + 5 a;"- 4 ) (2 z 3 - z 2 + x) = ? 9. (x 5 - i/ 6 ) ^- (x - y) = ? 10. (6 x m ~ n+z x m ~ n+1 22 x m ~ n 19 x" 1 """ 1 4 z m ~ n ~ 2 ) -^ (3 x 3 -" + 4 x 2 -" + or") = ? 3. Some important cases of factoring depend on the fol- lowing: 1. (a + &) + &) = (a + 6) 2 = a 2 + 2 ab + 6 2 . 2. (a - 6) (a - b) = (a - 6) 2 = a 2 - 2a6 + 6 2 . 3. (a + 6) (a - 6) = a 2 - 6 2 . 4. (a + x) (a + #) = a 2 + (x + y) a + xy. 5= (a + b) (a 2 - ab + & 2 ) = a 3 + 6 3 . 6. (a - 6) (a 2 + ab + 6 2 ) = a 3 - b 3 . 7. (a 6) 3 = a 3 3 a 2 6 + 3 a& 2 6 3 . 8. (a + &) n = a" + wa"" 1 6 + (n (n - 1)/1 2) a n ~ 2 6 2 + (n(n-l)(n-2)/1.2-3)a n - 3 & 3 + - + nab"- 1 + 6". 9. (a n - 6 n ) = (a - 6) (a"- 1 + a n ~ 2 & + a"- 3 b 2 + + ab n ~ 2 + &"- 1 )- 10. ax ay + bx by = a (x y) + b (x y) = (a + &) (x ?/). 11. (afec) 2 = a 2 + & 2 + c 2 Factor the following: 1. 20 db - 28 ad - 5 be + 7 cd. 2. 49 x 6 - 168 x 3 ?/ + 144 y 2 . REVIEW OF ALGEBRA 3. 6 xy + 16 z 2 - 9 y 2 - z 2 . 4. o 2 -2a6 + & 2 -c 2 . 5. a 2 - 10 a - 75. 6. a 2 -2a& + 6 2 -3a + 3& + 2. 7. x 2 + 2 xy - 99 i/ 2 . 8. 36 re 2 +12 a; -35. 9. z 3 -3z 2 + 3z-l. 10. a 3 -6a 2 + 18a-27. 11. 8x 3 -27y 3 . 12. x 3 -!. 13. a 9 -!. 14. a 2 w 2 + a + m. 15. (Sz 3 - 27) - (2z - 3) (4z 2 + 4z - 6). 16. x 3 + 2 a% + 2 z?/ 2 + y 3 . 17. (a 2 + 6 a + 8) 2 - 14 (a 2 + 6 a + 8) - 15. 18. x*-x 5 - 32 ^ + 32. 19. 18z 5 & + 24z 3 & 3 + 8z6 5 . 20. 27 tf - 108 x*y + 144 xy* - 64 j/. 21. 21 a 2 + 23 06 + 6 6 2 . 22. 10 a 2 -39 a + 14. 23. 95-14^-s 4 . 24. x + 512. 25. 5 x + 25 Z?/ 4. The lowest common multiple of two or more expressions is that expression of lowest degree that is exactly divisible by each of the given expressions. Thus: 1. a 2 6 3 c is the lowest common multiple of a 2 , a 2 6 3 and aWc. 2. y? -f- 6 3 is the lowest common multiple of x + 6 and x 2 - bx + fe 2 . Remark. In calculating the lowest common multiple of several expressions note that it must contain as factors every different factor of all the expressions. In case a factor "occurs more than once in any of the expressions it is to be taken to the highest power that it occurs in any of the expressions. THE HIGHEST COMMON FACTOR Find the lowest common multiple of the following: 1. 3z 3 -13z 2 + 23z-21 and 6x3 + a; 2 -44 a; + 21. (Try 3 x 7 as factor.) 2. z 2 + x 4 , 2 x z - 4 x and x 2 + 1. 3. a 2 + ab + ac + be and a 2 + 2 ab + 6 2 . 4. a 2 - 15 a + 50, a 2 + 2 a - 35 and a 2 - 3 a - 70. 5. z 2 + 5 x + 6, a? - 19 x - 30 and re 3 - 7 z 2 + 2 a; + 40. 6. a 3 + 6 a 2 + 11 a + 6 and a 4 + a 3 - 4 a 2 - 4 a. 7. 8a 2 -6a -9 and 6a 3 -7a 2 -7a + 6. 8. x 2 - 7a# + 12 y z , x z - Qxy + 8y 2 and z 2 - 6. The highest common factor of two or more expressions is that expression of highest degree that will be an exact divisor of the given expressions. Thus, abc is the highest common factor of o& 2 c 2 , a?bc, abc?. The highest common factor must contain as factors all the factors that are common to all the expressions. The highest common factor is usually obtained by factoring each of the given expressions and then taking each factor as many times as it is common to all the expressions. The product of these factors is the highest common factor. When the expressions are not easily factored a method due to Euclid * is useful. Let A and B be two expressions (or num- bers). Suppose (1) A = qB + R l} (2) B = q^ + R z , (3) Ri = qJh + R 3 , (4) R 2 = q3 R 3 . From these we have, since Rz = q^Ra, Rs is the highest factor of #2, Ri, R 3 . Hence B = R 3 (q&qa + q\+ qz). A = R 3 (qqiqtfs + qqi + qqs + q^z + 1). Therefore, the highest common factor of A, B is R 3 . This suggests the following rule: Divide A by B, call the quotient q and the remainder RI. Next, divide B by Ri, call the quotient q\ and the remainder Rz. Next, divide #1 by Rz, call the quotient q z and so on, con- tinuing to divide the last divisor by the last remainder, until a * See Geometry. This method may be omitted if desired. 6 REVIEW OF ALGEBRA remainder is obtained. The last divisor used is the highest common factor of A, B. If it becomes evident during the above process that the final division cannot give a remainder 0, then A, B have no common factor different from unity. Remark. During the process outlined above we may, when necessary or convenient, divide or multiply any of the expres- sions A, B, Ri, R 2 , . . . by any number not a common factor of A, B without affecting the highest common factor. If a factor be used which is common to A, B, we must account for it in making up the highest common factor of A, B. The following example will illustrate the process: Find the highest common factor of Gz 3 + 7x 2 y -3xy 2 and 4x*y + 8xY -3xy* - 9y*. First we take x out of the first expression and y from the second. We then have 6x z + 7xy -Zy* and 4s 3 + 8x*y - 3xy z - Qy*. Since 4 x 3 is not divisible by 6 x 2 we multiply the second expres- sion by 3. We have then 6 x z + 7 xy - 3 y z )l2 x 3 + 24 x*y - 9 xy* - 27 j/(2 x + 5 y -Qxy z !Qx z y-Zxy*-27y* Multiplying by 3, 3 30x 2 y + 35 sy 2 - 15 y 3 Divide by -22 y\ (-44xy* - 66 y 3 ) ^ (-22 y 2 ) = 2x 2 x + 3 y) 6 x 2 + 7 xy - 3 y 2 (3 z - y 6 s 2 + 9 xy Therefore 2 x + 3 y is the highest common factor of the given expressions. MANIPULATION OF FRACTIONAL EXPRESSIONS Find the highest common factor of: 1. 16, 72, 144. 2. 1728, 576. 3. a 2 - 5 ab + 4 6 2 and a 3 - 5 o 2 6 + 4 6 s . 4. 2 a 2 - 5 a + 2 and 12 a 3 - 8 a 2 - 3 a + 2. 4 5. z 2 - 1, x 3 + 1 and a; + 1. 6. 10 (a: + I) 3 and 4 (3 + I) 2 (a: - 1). 7. z 4 - 3z 2 - 28, x 4 - 16 and z 3 + z 2 + 4z + 4. 6. All the preceding processes may be used in the reduction and manipulation of fractional expressions. 1. (a 2 - 8 06 + 7 6 2 )/(a 2 - 3 ab + 2 6 2 ) to lowest terms. 2. (x 6 if) (x y)/((x* y 3 ) (x* t/ 4 )) to lowest terms. 3. (a + 6) ((a + 6) 2 - c 2 )/(4 aV - (a 2 - 6 2 - c 2 ) 2 ) to lowest terms. 4. (3-x)/(l-3x) -"(3+a;)/(l+3a;) - (l-16z)/(9z 2 -!), combine and reduce. 5. x/(x 2y) y/(2 y x) (x y^/(x 2 4 7/ 2 ), combine and reduce. 6. x/(x + 2) -3/(a; - 4) + 3/(x - 6) - l/(z - 8), combine and reduce. 7. (x 2 - yz)/((x + y)(x + z)) + (i/ 2 - zx)/((y + z) (y + x)) + (z 2 xy)/((z + x) (2 + t/)), combine and reduce. 8. 1/(1 - x) -JL/0 +s)/l/0 - z 2 ) - 1/(1 + x 2 ), combine and reduce. 9. (2-x-(6x combine and reduce. 10. (27z 3 combine and reduce. 11. 1 _ 3 + 1*3 4 + 3-5 (Begin at the bottom; combine and reduce.) 5-7 12. a+1 a+1 a+1 a+1 a 8 REVIEW OF ALGEBRA 7. To solve an equation is to find a value of some letter in the equation which will, when substituted for that letter, verify or satisfy the equation. The value of a letter (unknown quantity) in an equation which will satisfy the equation is called a root of the equation. Then, to solve an equation is to find its root or roots. It is assumed that the student knows how to solve simple equations, including equations containing fractions. The following exercises will afford review and extension of that knowledge. 1. (6z+ 1)/15 - (2x -4)/(7z - 16) = (2 x - l)/5. Hint. Multiply each member by the lowest common multiple of the denominators and solve. That is, clear of fractions and solve the equations for x. 2. l/(x - 2) - l/(x - 3) = l/(x - 4) - l/(x - 5). 3. (x + 2)/(* - 3) + (x - 3)/(* + 4) - (x + 4)/(z + 2) = 3. 4. (2x- l)/(2 x - 3) - (x 2 - x)/(x* + 4) = 2. 5. (x - !)/( - '2) + (x - 5)/(z - 4) + (x - 7)/(z - 6) + (x - 7)/(z - 8) = 4. 6. a/(x a) b/(x 6) = (a b)/(x c), where a, 6, c are regarded as known. 7. bx/a - (a 2 + 6 2 )/a 2 = a 2 /6 2 - x (a - b}/b. 8. (x 3 + !)/(* + 1) - (x* - l)/(x - 1) = 20. Note. It is better in this case to reduce the fractions to lowest terms before solving. 9. (ax - a 2 )/(x - 6) + (bx - tf)/(x - a) = a + 6. 10. (m + ri)/(x + m n) 2 m/(x m -f- ri) + (m n)/ (x m ri) = 0. 8. A single equation is sufficient to determine the value of one unknown symbol in the equation. We are to think of an equation containing one unknown as a condition to be satisfied by the unknown. We may impose one independent or arbi- trary assumption on one unknown. This assumption may be put in the form of an equation. It is then a problem, more or A SINGLE EQUATION 9 less difficult, to determine the value of the unknown that will satisfy the condition. That is, we must find a root of the equation, or, as we say, solve the equation for the unknown. We may make two independent assumptions or impose two independent conditions on two unknowns simultaneously. These conditions may be put in the form of equations. We must then solve the equations to determine the unknowns. To do this we must derive from the two equations a single equation with only one unknown. This equation will when solved give the value of one unknown. By substituting this value for that unknown hi one of the original equations there will result an equation with the other unknown which may now be deter- mined. The process of deriving a single equation with one unknown from two equations each containing two unknowns is called elimination of an unknown or symbol. When three equations each containing three unknowns are given for solution, we must derive two equations each contain- ing the same two unknowns and from these finally derive a single equation with only one unknown. The last equation being solved gives the value of one unknown. This value when substituted in one of the two equations will give an equation from which another unknown can be determined. Having the values of two unknowns they may be substituted in one of the original equations and then from the resulting equation the value of the third unknown can be determined. In general, the process of deriving re 1 equations with re 1 unknowns from n equations with n unknowns is called elimination. The idea of elimination and substitution is very important. Elimination is carried out in elementary algebra by several methods. Experience only can teach the pupil the best method to use in a given case. The following three methods are the ones most used: 1. Combine two equations by addition or subtraction so as to cause one unknown to disappear. Often one or both equations must be multiplied or divided by some known number in order that addition or subtraction will cause an unknown to disappear. 10 REVIEW OF ALGEBRA 2. Solve one of two equations for one unknown in terms of the other unknowns and substitute this value in the other equa- tions. The resulting equations will contain one less unknown than before. 3. Solve each of two equations for the same unknown in terms of the other unknowns and place these two values equal to each other. The resulting equation will contain one less unknown than the two before. For other methods books on higher algebra or theory of equations must be consulted. 1. Eliminate c from ay = x c, 2 ay'y = 2 (x c). 2. Eliminate c from ax + by = c, a'x + b'y = 3 c. 3. Solve for x, y from ax + by = c, a'x + b'y = c' and save the result as a formula for solving Ex. 4. 4. Apply the results of Ex. 3, to solve 2 x + 3 y = 8, 5x-2y-3. 5. Solve 10/s - 9/y = 8, 8/x + 15 /y = 1. Note. Do not clear of fractions, but consider l/x, l/y as unknowns. 6. Solve 6 x - 4 y X z = 17, 9 x - 7 y - 16 z = 29, 10 a; - 5 y - 3 z = 23. 7. If I = Prt and A = P (1 + rt) and if A = 1250, r = 0.06, 7 = 250, find P, t. 8. Solve a\x + biy + c\z = d\, azx + b 2 y + c& = dz, a$x + b s y + c 3 z = d 3 and save the result as a formula to solve Ex. 9. 9. Apply the result of Ex. 8, to solve 3 x 2 y -}- z = 5, 2s + 3 y- 4 = 8, x-5y + 3z = 6. 10. Eliminate x from the equations xy = 12, x y = 1. 11. If Z = a + (n - 1) d, S = (a + 1) n/2, and if a = 4, /S = 60, I = 16, find n and d. 12. If z = 19, y = 1900; x = 25, y = 3230; re = 44, y = 9780 satisfy the equation y = ax 2 + bx + c, determine a, 6, c. 13. If a straight line passes through the points whose co- ordinates are x = 2,y = 3; x = 4, i/ = 6 and if the equation of the line is y = mx + b, find m, b. FORMULAS RELATING TO RADICALS 11 9. Formulas relating to radicals: All ordinary operations with radicals are based on the following: 2. a* 7 ' = 3. v^or 4. 5. \~\fa = "v/a. 6. 7. < 8. 6 9. vV" v'a"""' = a. 10. (V^ + Vb) (Va - Vb) = a - b. These formulas are extensions of the index laws. By means of formula 1 and the index laws in article 2 the remaining nine formulas are easily verified. Simplify the following: 1. VI21. Thus, Vl2l = VII 2 = 11. See formula (1). 2. A/625 x^y 4 ^. Thus, v/625 x l2 yW = ^^yz 2 = x >/5^. 3. V480; VI (use formulas 8, 9). Thus, V| = \/ x 4. ^81 a 4 /16 be 2 . 5. v/320 - V^135 + ^625. 6. Vf| + V|o + \/|. Reduce to common index: 7. V3, v/5, v^. Use formula (3), reduce exponents to common denominator and pass back to radicals. * If o is negative and n an even integer, Va is said to be an imaginary quantity. Such quantities are written as follows: V^a- = V- 1 Vx = i Vx where x is positive. 12 REVIEW OF ALGEBRA 8. Vl2 + V75 + VT47 - Vi8. 9. \/2 a 2 6, v'G 6 2 c 5 , "v/14 c 4 a 7 , to common index. 10. Vl2 + \/3 - V45. 11. V3, "V/5, VTl, to common index. 12. V50 - ViJ - v^24 - \/ll. 13. ViTaF 2 V8 b*<? (to common index before multiplying). 14. V^-Vf-V^. 15. (Ve + VTo + vTi) -5- V2. 16. V5 - 2 \/2 . V5 - 2 \/2. 17. (x + y) V(x - y)/x + y - (x - y) V(x + y)/(x - y] 18. (V3 - \/2)/(V2 - V5), use formula (10) to remove radicals from the denominator. 19. (a+ V6)/(o- Vft). 20. (3 + V6)/(\/2 + V3). Solve the following equations and substitute the results in the original equation in each case. 21. x% = 4, raise both members to the fourth power. 22. (V2x l)^ = V3, raise both members to the sixth power. 23. Vx 4 + Vx 11 = 7, square both members, trans- pose, square again. 24. l/(VxT~i) - l/(Vx~^T) + 1/Vz 2 - 1 = 0, clear of fractions and proceed as in (23). 25. V/2 + \/3 + Vx = 2. 26. z - 7 - Vx 2 5 = 0, transpose x 7 before squaring. 27. x* - 7 - Vz 2 - 4 = 0, x 2 is to be considered the un- known at first. 10. Each student should be able to solve quadratic equa- tions quickly and accurately by some method. The theory ol equations of the second degree and of higher degrees will be A PAIR OF SIMULTANEOUS EQUATIONS 13 treated in a later chapter. Here will be given a formula which will answer all present needs. Suppose the quadratic equation has been reduced to the form ax 2 + bx + c = 0. Then the two roots are given by the formula, 2a A pair of simultaneous equations with two unknowns, where one equation is of the second degree and one of the first degree, may be solved by the second method of elimination, 8. Solve the first degree equation for one of the unknowns in terms of the other and substitute in the second degree equation. The resulting equation will be a quadratic with one unknown which can be solved by the above formula. The other unknown can then be found by substituting in the first degree equation. Solve the following equations: 1. z 2 -2z = 35. 2. z 2 -3s-l - A/3 = 0. 3. 2x/(x + 2) - (x + 2)/2z = 2. 4. 15z 2 -86z-64 = 0. 5. 4 x* 17 x 2 18 = 0, consider x 2 as the unknown at first. 6. 3** -4z* = 7. 7. a 2 - 1/x* = a 2 - I/a 2 . 8. V3 x + V2 x = \/5 - 2 x. 9. 10> \2x-Zy = 5. ( 1/z 3 + 1/0* = 1001/125. Divide first equation by second ( 1/x + l/y = 11/5. member by member and solve resulting equation with second. x y = 1. 13> \xy = (a 2 - 6 2 )/4. 14 REVIEW OF ALGEBRA 11. Inequalities. Often the most important lact about two numbers is that one is greater than or less than the other. To express such relations briefly a sign of inequality is used as follows: a > b means a is greater algebraically than b, or that b is less than a. If 6 < a then b < a means not less than), for we know that b > a if a > b. The following laws hold for inequalities as for equations: 1. Both sides of an inequality may be multiplied or divided by the same positive number without affecting the sense of the inequality. 2. Equal numbers may be added to or subtracted from both sides of an inequality. A term may be transposed. If both sides of an inequality be multiplied or divided by the same negative number the ense of the inequality is reversed and the vertex of the sign must be pointed in the opposite direction. 1. For what values of x is the expression 7 x 23/3 < 2 z/3 + 5? Multiply both sides by 3, 21 x - 23 < 2 x + 15. Transpose, 19 re < 38. Divide by 19. x < 2. Therefore the inequality holds for all values of x less than 2. 2. For what values of x is (x - 1) (x - 2) (x - 3) < (x - 5) - 6) (x - 7). 3. Show that for all values of x, 9 x 2 + 25 < 30 x. 4. For what values of x is 4 x 2 4 x 3 < 0. First determine for what values of x the expression is 0. Then by inspection determine the ones that satisfy the in- equality. 5. Show a/b + 'b/a > 2, for all positive values of a, b and 6 7* a. 6. Show (a + 6) (a 3 + 6 3 ) > (a 2 - 6 2 ) 2 . 7. For what values of x is x 7 > 3 x/2 8. 8. For what values of x is (a; - 1) (x - 3) (x - 6) > 0. By choosing values of x and determining the signs of the factors the answer can be deduced without calculating. THE BINOMIAL FORMULA 15 12. The binomial formula is of frequent use in mathematics. Let us assume: (1) ( n(n-l)(n 2). n (n - 1) (n - 2) . . . (n - r + 2) a n - f + 1 fr- 1 1 2 3 ... (r - 1) 1-2- 3 . . . (r-l)r H + 6 n . Multiplying both sides by a + 6 and carrying out the actual multiplication, (2) (a + b) n+1 = a n+1 + (n + 1) a n b + ^-= . -|_ . . . (n + l)n (n 1) . . . (n r + 3) a n ~ r+2 b f - 1 1-2 ... (r-1) I (n-rL)n(nL). . . (n y+2) a n o' , , +1 1.2. . .r It is seen that equation (2) is exactly what equation (1) would become if in the latter n + 1 is put in place of n. By actual trial it is known that (1) holds for n 2. Then (2) gives immediately the value of (a + 6) 3 . Now in (1) put n = 3 and (2) gives the value of (a + 6) 4 . This process may be continued indefinitely. It follows that (1) and (2) hold for any and all positive integral values of n. It will be assumed here and proved later that (1), (2) hold for fractional and negative values of n. 1. Find without multiplying the 5th power of 1 x. 2. Find without multiplying the \ power of 1 x to 4 terms. 3. Find without multiplying the \ power of 1 # to 4 terms. 4. Find without multiplying the 1 power of 1 x to 4 terms. 5. Find without multiplying the \ power of 1 x to 4 terms. CHAPTER II. GEOMETRICAL THEOREMS AND FORMULAS* 13. 1. A circle of given radius can be described about any point as a center. 2. In the same, or in equal circles, equal central angles inter- cept equal arcs. 3. The perimeters of inscribed polygons are less than the circle and the perimeters of circumscribed polygons are greater than the circle. 4. If two straight lines intersect, the vertical angles formed are equal. 5. If two parallel lines are cut by a transversal, the alternate interior angles formed are equal. 6. If two parallel lines are cut by a transversal, the corre- sponding angles formed are equal. 7. Angles having their sides parallel each to each and extend- ing in the same direction or in opposite directions from the vertex are equal. If one pair of parallel sides extend in the same direction and the other pair in the opposite direction the angles are supplementary. 8. Angles having their sides perpendicular each to each, and both acute or both obtuse, are equal. 9. The sum of all the angles of a triangle is equal to a straight angle. 10. In any triangle there must be at least two acute angles. 11. In any isosceles triangle, the angles opposite the equal sides are equal. 12. An equilateral triangle is equiangular. 13. In an isosceles triangle the bisector of the vertical angle is the perpendicular bisector of the base. * These are given chiefly for reference use. 16 GEOMETRICAL THEOREMS AND FORMULAS 17 14. If two sides and the included angle of one triangle are equal to the corresponding parts of another, the triangles are congruent. 15. If two angles and a side of one triangle are equal to the same parts of another, the triangles are congruent. 16. If three sides of a triangle are equal to three sides of another, the triangles are congruent. 17. Two lines perpendicular to two intersecting lines, re- spectively, must meet. 18. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. 19. The diagonals of a parallelogram bisect each other. 20. If the diagonals of a parallelogram are equal, the figure is a rectangle. 21. The diagonals of a rhombus bisect each other at right angles. 22. The sum of the interior angles of a convex polygon of n sides is n 2 straight angles. 23. The sum of the exterior angles of a convex polygon is two straight angles. 24. Every point on the perpendicular bisector of a sect is equidistant from the ends of the sect. (A sect is a segment of a straight line.) 25. Any straight line parallel to a side of a triangle forms with the other two sides a triangle similar to the first. (The two sides produced if necessary.) 26. A line parallel to one side of a triangle divides the other two sides in the same ratio. 27. Homologous sides of similar triangles have the same ratio. 28. In any two similar polygons (1) the homologous angles are equal; (2) the homologous sides are proportional. 29. In any two similar polygons the ratio of any two homol- ogous lines is equal to the ratio of any other homologous lines and equal to the ratio of the perimeters of the polygons. 30. The longer side of a triangle is opposite the greater angle. 18 GEOMETRICAL THEOREMS AND FORMULAS 31. In any triangle any side is less than the sum of the other two sides and greater than their difference. 32. The diameter of a circle is twice its radius. 33. A circle is determined by (a) the center and radius; (6) center and diameter; (c) three points not in a straight line. 34. A perpendicular bisector of a chord passes through the center of the circle. 35. A straight line perpendicular to a radius at its extremity is tangent to the circle. 36. Equal chords are equally distant from the center of the circle. 37. Two angles at the center have the same ratio as their in- tercepted arcs. 38. The area of a rectangle or a parallelogram equals the product of the base by the altitude. 39. The area of a triangle equals the product of the base by half the altitude. 40. The area of a circle equals IT times the square of its radius, (ur 2 ). TT = 3.1416 approximately. 41. The area of a sphere equals 4?r times the square of its radius, (4 irr z ). 42. The sum of the squares of the legs of a right triangle equals the square of the hypotenuse. 43. The altitude of a right triangle from the right angle is a mean proportional between the segments into which it divides the hypotenuse. 44. The volume of a prism or cylinder equals the area of its base times its altitude. 45. The areas of similar polygons have the same ratio as the squares on any two homologous lines. 46. Area of triangle whose sides are a, b, c, is given by A = Vs (s a) (s 6) (s c), when 2s = a + 6 + c. 47. Area of sector of circle A = ~ GEOMETRICAL THEOREMS AND FORMULAS 19 48. Area of segment of sphere A = 2irrh where h is altitude of segment. 49. Area of curved surface of cylinder, A = 2 irrh. 50. Area of curved surface of cone, A = 2 irr ~ , where s is t slant height. 51. Area of a segment (of one base) of a circle is the area of the sector subtended by the arc of the segment minus the area of the triangle whose base is the chord of the segment and vertex the center of the circle. CHAPTER III METHODS OF CALCULATION 14. The need of numbers and number calculations arose quite naturally in race development. Recognition of number and geometric form appears not to be peculiar to the human race. There is good reason for believing that many of the lower races of animals have the ability to count and to recognize space forms and magnitudes. A few of the conditions in human affairs that called forth methods of counting, measuring and reckoning are enumerated below. 1. Numbering and comparing groups of objects. 2. Barter and trade of primitive commerce. 3. Census and tribal statistics. 4. Calendar and time measurements. 5. Land measurements. 6. Determination of weights and measures. 7. Taxation. 8. Modern commerce. 9. Engineering and exact science. 10. Personal and corporation accounting. 11. Insurance, savings accounts and investments. 12. Statistical and social science. 15. Important discoveries in the art of calculating. (a) -Hindu notation: We owe our present method of writing ordinary numbers to the Hindus. The important principles of this notation and the ones that make it superior to all others, to date, are its position value and zero. This notation probably antedates the Hindus but it was through them that it was transmitted to the western nations. 20 MECHANICAL CONTRIVANCES 21 (6) Decimal fractions were given to the world about the end of the sixteenth century by Simon Stevins of Bruges in Belgium. Fractions had been a hard problem for the race. For a long time all fractions were expressed as the sums of unit fractions. Thus 7/12 was regarded as 1/2 + 1/12 and similarly for other fractions. Other methods were developed gradually. The discovery that fractions could be incorporated into the number system in decimal form as an extension of the Hindu notation was a real contribution not only to mathematics but to civili- zation. (c) Early in the seventeenth century the discovery of loga- rithms gave the computer and mathematician another power- ful instrument. It was the crowning achievement in numbers. Coming at a time when Kepler was working on his planetary theory and immediately following the computation of trigono- metric tables by the Germans, the discovery of logarithms was of great value. The names of Napier, the inventor, and Briggs, the collaborator and editor, of the tables have been made immortal by these contributions. Laplace remarked that the discovery of logarithms would double the life of the astronomer by shortening the labor of his calculations. 16. Many mechanical contrivances have been and are still in use for shortening the labor of computing. In passing we may mention : (a) The abacus, an ancient instrument, still in use in some countries and used in our schools under the name, numeral frame; (6) the slide rule, which is based on the idea of logarithms. It is convenient and adequate for many purposes in applied science and in commerce; (c) the calculating machine, such as the Burroughs, does all the ordinary arithmetical calculations with a speed and accuracy that justifies its use in banks and mercantile houses where much computing is done; (d) geometrical methods of performing the arithmetical opera- tions were in possession of the Greeks. In certain branches of engineering, geometric methods are now used with great efficiency. They depend on the principle of the proportionality of the sides of similar triangles, (e) General reckoning tables, 22 METHODS OF CALCULATION containing products, quotients, powers, roots and reciprocals of numbers may be obtained by those who desire to use them. 17. Graphic or geometric representation of numbers. - A number is graphically represented by a straight line of length proportional to the magnitude of the number. The factor of proportionality determines the scale of the representation. Thus, if a line 1" * long is to represent the number 50, the scale is 50 to 1 in inches. If the number 275 is to be represented by a line not over 4" long, we divide 275 by 4 and obtain 69, nearly. It would, therefore, be safe to use a scale of 70 to 1 or any larger number than 70 to 1. It must be remembered that the smaller the scale (larger ratio of proportionality), the more difficult it is to estimate small numbers or odd units. On the other hand the larger the scale (smaller ratio of proportionality), the larger the drawing and the larger the paper or other surface required. The scale is to be determined by convenience and by the degree of accuracy demanded. Very large drawings are sometimes nec- essary. Sometimes very small drawings will do. 1. (a) Lay off 350 to the scale of 50 to 1"; (6) lay off 350 to the scale of 100 to 1". 2. Measure the lines below to the scale of 50 to 1" and deter- mine the numbers they represent. A FIG. 1. 18. Arithmetical operations. (1) To add a number 6 to a number a: Let 01 be the unit length. Suppose a = OA, b = AB. Obviously OB = OA + AB represents the sum of a and 6, to the unit 01. B FIG. 2. * The notation, 1", means 1 inch; 1' means 1 foot. ARITHMETICAL OPERATIONS 23 To subtract b from a lay off 6 from A toward B'. Then a - b = OA - AB = OB'. (2) To multiply a number a by a number 6. Let 01 be the unit. Lay off a = OA. From A draw a line AC making any convenient angle with OA. Lay ofi IB parallel to AC and equal to b. Draw OB and produce it until it intersects AC at C, say. Then AC represents the product ab. That is, c = ab. For, by the similar triangles OAC, 01B, 01 : IB : : OA : AC or 01 - AC = IB - OA and AC = IB OA, c = ba, since 01 = 1, the unit of measure. FIG. 3. (3) To divide a number c by a number a. In this problem use the diagram of the preceding problem and solve the last equation for b. It is seen that if a is the divisor and c the divi- dend, the quotient is 6. The reciprocal of a number can be found as a special case where the dividend is 1. In using this method for finding 24 METHODS OF CALCULATION reciprocals, it is advisable to employ a larger scale for the dividend than for the divisor. The square of a number can be found as a special case of problem (2) where a = b. (4) To find the square root of a number a: Evidently a = a 1 and Va = VoTI. Make BC = a and AB = 1. Draw on AC as a diameter, a circumference, AKC. At B erect a per- pendicular meeting the circumference at D. Then BD Z = AB - BC by geometry. Hence Va = 5D. >c 1. Find the product of 35 X 73; 18 X 93; 36 X 13 by the geometric method. 2. Find the quotient of 125/16; 39/4; 50/13 by the geo- metric method. 3. Find the reciprocals of all integers from 2 to 10, inclusive, by the geometric method. 4. By the geometric method find Vl2; V27; Vl8; V56. 5. By the geometric method find (35 X 12)/16; 18x27/13. 19. Idea of logarithms. The student should now recall the index laws, see 2, Chap. I. All positive numbers can be regarded as powers of some positive number, not unity. It is, of course, true tkat most such powers cannot be exactly expressed, but close approximations can be determined and exactly expressed. This is because these indices are not rational numbers. A rational number can be expressed as the quotient of two integers. RULES 25 What power of 2 is 8? What power of 2 is 10? What power of 10 is 10? What power of 10 is 100? What power of 10 is 1/10? 20. Definition of the logarithm of a number. If y = a x , x is called the logarithm of y to the base a. Briefly we may write x = log a y. y = a x implies x = log a y and conversely. From this definition it follows that logiolO = 1, since 10 = 10 1 , logio 100 = 2, since 100 = 10 2 , logic 1/100 = -2, since 1/100 = 10~ 2 . 21. Rules for calculating with logarithms are derived directly from the index laws. Thus, 10 2 10 1 = 10 2+1 = 10 3 = 1000. That is; (1) logic (10 2 10) = log 1000 = 3 = 2 + 1 = logio 100 + logio 10. Again : (2) logio (1000 ^-10)= logio 100 = 2 = 3-l=logio 1000-logio 10. Generalized, (1) and (2) give the rules for multiplication and division by means of logarithms. Let x and y be any two positive numbers and p their product. Then, (!') loga p = loga x + logo y. If q is the quotient x/y, then (2') log a q = loga X - loga y. Consider (3) y = x n , where n is any number. Then, by 20 loga y = logo yory = a 10 **, x = a 10 *- 1 , x n = (a 108 * 1 )" = a* 10 *- 1 . Therefore, (3') loga y = loga x n = n log a x. This result shows how to use logarithms to raise a number to any power. 26 METHODS OF CALCULATION Now consider: i (4) y = Vz = x m . We can write: L I (40 10g a y = bga OT = loga X. Ub This result shows how to use logarithms to find the roots of numbers. The foregoing equations (I'), (2'), (30, (40 m ay now be translated into rules. 1. The logarithm of the product of two or more numbers is the sum of the logarithms of the numbers. 2. The logarithm of the quotient of one number divided by another is the logarithm of the dividend minus the logarithm of the divisor. 3. The logarithm of the nth power of a number is n tunes the logarithm of the number. 4. The logarithm of the nth root of a number is the logarithm of the number divided by n. 5. The logarithm of the reciprocal of a number is the negative of the logarithm of the number. (See co-logarithm below.) It may be more convenient to use addition when dividing with logarithms. To do this the logarithm may be subtracted from or 10 10, or 20 20, etc. Thus instead of subtract- ing log n = 3.37581, it may be more convenient to use addition. This is easily done by first subtracting 3.37581 from 10 10. Thus we may add 6.62419 10. The advantage of this method lies in the fact that it is easy to make the above subtraction directly as the logarithm is taken from the table by subtracting each figure, beginning at the left, from 9 and the last figure on the right from 10. The result of this subtraction is called the co-logarithm of the number or arithmetic complement of the logarithm. 22. From 19 and 20 it is evident that when 10 is the base of the logarithms, the following statements hold : RULES 27 1. Any number between 1 and 10 has a proper fraction for its logarithm. 2. Any number between 10 and 100 has 1 plus a proper fraction for its logarithm. 3. Any number between 100 and 1000 has 2 plus a proper fraction for its logarithm. 4. Any number between and 1 has a negative logarithm. It is seen that logarithms of numbers consist of two parts, an integer or zero and a fraction. The integer part is called the characteristic. The fractional part is called the mantissa of the logarithm. For convenience in calculating, a logarithm should be written so the part on left of decimal point is positive. The mantissa is always positive. Thus, 2.34567 (where 2 shows that the char- acteristic, only, is negative) may be expressed as 8.34567 10 where the part of the characteristic on left of the decimal point is positive. Consider: log 273 = 2 plus a proper fraction. log 27.3 = 1 plus the same proper fraction. log 2.73 = plus the same proper fraction. log 0.273 = I plus the same proper fraction. log 0.0273 = 2 plus the same proper fraction. These results are evident from the fact that, when the decimal point is moved one place to the left, the number is divided by 10 and exactly 1 is subtracted from its logarithm, the mantissa remaining the same. It is easy to see that the characteristic depends on the position of the decimal point in the number, while the mantissa depends on the sequence of digits in the number. It will be noticed that with a number = 1 the char- acteristic is an integer one less than the number of significant figures of the number to the left of the decimal point. This rule is general if the number of O's immediately to the right of the decimal point be considered as a negative number of figures on the left. Note that the above example verifies this statement. 28 METHODS OF CALCULATION What is the characteristic of the logarithm of each of the following numbers: 3.047, 37.56,0.000842, 1.0045, 67,543,0.43? 23. In order to use logarithms in calculating it is necessary to have a table of logarithms of all numbers used in the cal- culations. The mantissas only are given in the table. The characteristic must be supplied in every case according to the above rule. To construct a table of logarithms involves great labor. It is beyond the scope of this course to explain the method of cal- culating such tables. For this information the student is referred to works on higher algebra or calculus. A series from which logarithms may be calculated will be given in Chap. XVI. A four place table of logarithms of numbers with explanations is to be found in the back of the text. Solve the following by the use of logarithms. 1. 79 470 0.982. By the rule for multiplication of 21 write log (79 470 0.982) = log.79 + log 470 + log 0.982 = 1.8976 +2.6721 +9.9921 -10 = 1.9921 = 14.5618 - 10 = 4.5618 = log 3646. 2. 9503 0.7095 = ? 5. (2.58S) 5 = ? 3. 8075 -4- 364.9 = ? 6. (0.57)- 4 = ? 4. (-0.643) . 0.7564 = ? 7. ^-0.3089 = ? 8. (0.000684)* = ? 9. V943 (-7298)7(0.00006. (-9~9)) = ? 10. (\/0.0476 ' ^222)7 ^5059 0.0884 = ? 24. An exponential equation is one in which the unknown quantity occurs as the exponent of some known number in the equation. Such equations can often be easily solved by the use of logarithms. AN EXPONENTIAL EQUATION 29 Find the value of x in the equation 13 2 * = 14 r+1 . Taking logarithms of both sides, 2zlogl3 = (x+l)log!4 and , log 14 2 log 13 -log 14* Substituting the logarithms of 13 and 14 gives x = 1 .06. Use logarithms in solving the following problems. 1. Given A = P (1.0 r)', I = A P, find the compound interest of $250 at 5 per cent for 25 years. t (P = principal, r = rate, t = time, A = amount.) Note. .Or may be written r/100 if preferred. 2. In how many years will $5000 amount to $6000, com- pounded annually at 6 per cent? Given A = P (1.0 r)*. 3. Find the value of x in 5*+ 5 = 8 I+1 . 4. Find the value of k in P s = P<fT k *, if P = 14.72, z = 1122, P, = 14.11, e = 2.718. 5. How many gallons in a cylindrical tank 3' in diameter and 6' 8" high? Note. 6' = 6 ft.; 8" = 8 in. This notation will be used from here on. 6. If c = (Ld/fl6) (26 2 - n 2 ), find c when L = 1650, b = 500, n = 25, R = 1000, d = 0.75. 7. Compute the simple interest of $135.70 at 3| per cent for 12$ yrs. (7 = PH.) 8. In the triangle ABC, a = 175, 6 = 225, c = 190. Find the area. (A = Vs(s a) (s 6) (s c) and s = (a + b + c)/2. 9. Find the cost of covering a floor like the figure at $1.75 per sq. yd. < t r,o' ^ ^s *bo I ..11'. > FIG. 5. 30 METHODS OF CALCULATION 10. Find the number of gallons of paint required to paint a cylindrical tank 10' in diameter and 40' long (curved surface and both ends to be painted), if 1 gal. paint is sufficient to paint 100 sq. ft. of surface. 11. A grindstone will stand a rim speed of 2500' per min. How many r.p.m. (revolutions per min.) will a stone 42" in diameter stand? 12. If the cutting speed of a lathe is 40' per min., how many r.p.m. must a lathe have to give the desired speed for cutting a piece 2" in diameter? 13. Find the number of barrels capacity of a rectangular cistern 6' X 8' X 10'. (7.48 gal. = 1 cu. ft.) 14. A pie is 10" in diameter. It is cut in 6 equal sectors about the center. What is the area of the upper surface of each piece? 15. Find the value of a bin of wheat 8' 8" X 10' 6" X 4' 3" at $1.95 per bu. 25. The logarithmic scale. If distances proportional to the logarithms of numbers are laid off from one end of a straight line segment AB, the segment so divided is a logarithmic scale. 1, 2 3 456789 10 A B' FIG. 6. If a duplicate scale A 'B' is laid along side AB so as to slide parallel to itself, we can use these scales to perform the opera- tions of multiplication and division of numbers. For example, to multiply 25 X 30, set A' at 25 (between 2 and 3) on AB. Run over on A 'B' to 30 (at 3 on A 'B') and on A B opposite this read 75 (between 7 and 8) on AB. This gives the figures of the product. Knowing the orders of the numbers we know the product is 750. Since the scales are proportional to the logarithms of numbers, what has been done is to add the logarithm of 30 to the loga- rithm of 25 to obtain the logarithm of 750. An instrument operating in this way is called a slide rule. Directions for use 31 generally accompany the rule. The student should now obtain a slide rule and learn to use it as an economy and as a check in his calculations. Commercial and engineering concerns employ the slide rule for certain types of work with great advantage. 26. For convenience of reference, rules for the ordinary Manheim slide rule are appended here. Multiplication. Set the index of C scale over multiplicand on D scale. Under the multiplier on C scale read product on D scale. Division. Above dividend on D scale set divisor on C scale. Under index of C scale read quotient on D scale. Proportion. Set first term on C scale over second term on D scale. Under third term on C scale read fourth term on D scale. Squares. Set the runner on the number on D scale. Read square under runner on A scale. Square root. If number has an odd number of integral digits use left half of A scale; if an even number of integral digits use right half of A scale. Set runner over number on A scale. Read square root under runner on D scale. Cubes. Set index of B scale under number on A scale. Read cube on A scale over number on C scale. Cube root. Under number on A scale move slide until number on A scale above index of B scale is same as appears on C scale under given number on A scale. Exercise. Solve all problems of 18 by use of the slide rule. 26a. Methods of Interpolation. Immediately preceding the tables in the back of the text is an explanation of the use of those tables. A number of problems are worked which illustrate simple interpolation. Below are given problems which illustrate "double in- terpolation." Other types of interpolation are given in Chapters IV and VIII. Following is a section of a table given in the (J. S. Weather Bureau Bulletin No. 235. A similar table is used in connec- tion with work in artillery and in the navy. 32 TABLE. RELATIVE HUMIDITY, PER CENT FAHRENHEIT TEMPERATURES Pressure = 29.0 Inches Depression of wet-bulb thermometer (t-t r ). Air temp. /. 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 40 96 92 88 84 80 76 72 68 64 61 57 53 49 46 42 38 35 31 27 23 41 96 92 88 84 SO 77 73 69 65 62 58 54 50 47 43 40 36 33 29 26 42 96 92 88 85 81 77 73 70 66 62 59 55 51 48 45 41 38 34 31 28 43 96 92 88 85 81 78 74 70 67 63 60 56 52 49 46 43 39 36 32 29 44 96 93 89 85 82 78 74 71 68 64 61 57 54 51 47 44 40 37 34 31 45 96 93 89 86 82 79 75 71 68 65 61 58 55 52 48 45 42 39 36 33 Example: Air temperature t = 42.7. Reading of wet-bulb thermometer t' = 34.5. Depression of wet-bulb thermometer (t t'} = 8.2. Determine the relative humidity from above table. Tabulated values are : 8.0 8.5 42 41 38 43 43 39 To a difference of 1 in air temperature (43-42) corresponds a difference of 2 per cent in the relative humidity (43-41) when depression is 8.0. Therefore, a change of 0.7 in air temperature (42.7 42) causes a change of 0.7 X 2 per cent = 1.4 per cent in relative humidity. Similarly, for a depres- sion 8.5, a difference of 1 in air temperature causes a differ- ence of 1 per cent in relative humidity, and, therefore, a change of 0.7 causes a change of 0.7 per cent in the relative humidity. Our table may now be enlarged to 8.0 8.5 42 41 38 42.7 42.4 38.7 43 43 39 METHODS OF INTERPOLATION 33 When the depression is 8.2, the relative humidity by the usual method of interpolation is: Difference of .5 depression causes a difference of 3.7 per cent in relative humidity (42.4 38.7). Therefore, difference of 0.2 depression causes a difference of 1.5 per cent in relative humidity. The required humidity is therefore (42.4 per cent - 1.5 per cent) = 40.9 per cent. 1. Work the above illustrative problem by changing the order of the method, that is, interpolate for temperature 42 and depression 8.2, which lies between 8.0 and 8.5. Then do likewise for temperature 43 and so forth. 2. If air temperature, t = 44.3, Reading of wet-bulb thermometer t' = 34.5, Depression of wet-bulb thermometer (t t') = 9.8, Determine the per cent humidity. 3. Temperature is 41.8, Depression of wet-bulb thermometer (t t'} 2.4, Determine the per cent humidity. The motion of a projectile is retarded by the resistance of the air. The amount of retardation due to this cause depends upon the temperature and barometric pressure. Following is a portion of a ballistic table from which is determined an "atmospheric factor" that enters into the computation of retardation 1 BALLISTIC TABLE Temperature of air Fahrenheit degrees. 30 31 32 33 34 35 36 37 38 28" 0.994 0.996 0.998 1.000 1.003 1.005 1.007 1.009 1.011 29 0.960 0.962 0.964 0.966 0.968 0.970 0.972 0.974 0.976 30 0.928 0.930 0.932 0.934 0.936 0.938 0.940 0.943 0.945 31 0.898 0.899 0.902 0.903 0.906 0.907 0.909 0.911 0.913 1. Given the temperature t = 32 and the barometric pres- sure p = 29", then the corresponding atmospheric factor is 0.964. Verify by use of the table. 34 METHODS OF CALCULATION 2. Given the temperature t 31.8 and the barometric pressure p = 28.4". Find the corresponding atmospheric factor. Note. Carry out the interpolations in the same manner as was done in the illustrative problem above on humidity of the air. 3. If t = 31 and p = 30.8", find the corresponding atmos- pheric factor. 4. If t = 36.6 and p = 30.2", find the atmospheric factor. SUPPLEMENTARY EXERCISES 1. The diagonal of a square is 73.84'. Find the length of a side. 2. The circumference of a circle is 63.24'. Find the radius. (Use 3.142 for TT.) 3. If the area of a circle is given by the formula A = 0.7854 d 2 , find the radius of a circle whose area is 842.3 sq. ft. 4. If 1" = 2.540 cm., how many centimeters in 8 T y? 5. If the barometer reading is 29.92", what will a barometer read that is graduated in millimeters? If the barometer reading is 75.46 cm., what will a barometer read that is graduated in inches? 6. To find the reciprocal of a number by the use of logarithms: Sub- tract the mantissa of the log of the number from 1, add 1 to its character- istic and change the sign. The number whose log : s this result is the reciprocal of the given number. Use this rule in finding the reciprocal of 543. Find log 1/543 by use of a co-logarithm. How do the two methods for finding the reciprocal of a number differ ? 7. Find the reciprocal of 0.00635 by the two methods of the above problem. 8. If 1 cu. ft. of water weighs 62.45 Ibs., what is the pressure per sq. in. of a column of water 1 ft. high? 9. The volume of a sphere v = 4/3 irr 3 . Show that r = ( j j and find r in cm. when v = 37.45 cu. in. 10. The time required for a simple pendulum to make a single oscilla- tion when the angle through which it swings is small, is expressed by t = IT VT/0, where I represents the length of the pendulum and g represents the acceleration of gravity. What is the length of a pendulum that vi- brates seconds when g = 979.4 cm. /sec 2 ? 11. The equation of motion of a body falling from rest under the action of gravity is s = f gt 2 where s = distance and t = tune. METHODS OF INTERPOLATION 35 (a) If g = 32.16 ft./sec 2 and s = 164.8 ft., find t. (b) If g = 980.3 cm./sec 2 and s = 50.23 m., find t. Note that g is expressed in centimeters and s in meters. Must this be taken into con- sideration when computing t ? 12. If a 250 lb. charge of a certain lot of nitro-cellulose powder at 70 F. gives a muzzle velocity of 2275 ft./sec., what should be the weight of a charge to give a muzzle velocity of 2750 ft./sec.? Muzzle velocity and V f w\ v weight of charge are connected by the formula =~- = f 1 . y = 1.2 for nitro-cellulose powder. 13. Using same formula as in problem (12) find the muzzle velocity for a charge of 275 Ibs. where the velocity is 2150 ft./sec. at 30 F., for a charge of 240 Ibs. of nitro-glycerine powder, y = 0.8 for nitro-glycerine powder. 14. The following formulas are used for range correction in gunnery: r - f c f -1 i 2W * T * wl Jw ' t-'j Jw * "I Y Find d when T = 36.60, W x = 25.00, X = 54,000, and C = 10.49. 16. Given (J) n = J; find n. Note. Take the reciprocal of each member. 16. Given (l/2.641) n = 1/(159.4); find n. 17. Given (l/(0.0649) n = 1/2.54; find n. nl rVA*"" 1 /Po\ n ^ I = ( TT J ; find n. (F,\ n ~i /P lj = ( L? Note. Raise both members of the equation to the power n/(n 1). 19. In problem 18 find n when Vi = 0.5, V z = 6, PI = 250 and P 2 = 7.71. CHAPTER IV GRAPHIC REPRESENTATION 27. Modern life and scientific investigations make it neces- sary to use statistics and to draw conclusions from statistical records. These records usually consist of tabulations of measurements in the form of columns of figures. It is not always easy to get all the information contained in a table of statistics without some kind of pictorial aid. When a table of numbers can be put in a form that appeals through the eye to our space notions it is easier to understand and to draw con- clusions from such a table. Various representations to this end are given the name, graphic representations. Curves, geometric diagrams, pic- tures, etc., are frequently used for this purpose. Of the above representations curves are most widely used. In the sequel our study of graphic representation will be limited to the method of curves. The method is best explained by the use of some simple examples. 1. Suppose a man walks for several hours on a winding path and keeps a record of the distance traveled each hour. Let the annexed diagram represent the path and the points reached at the end of each hour. 1st hr., | mi. 6th hr., 1 mi. 2d hr., If mi. 7th hr., f mi. 3rd hr., 2 mi. 8th hr., If mi. 4th hr., f mi. 9th hr., \ mi. 5th hr., \ mi. 10th hr., 2 mi. To construct the graphic representation of this record proceed as follows: Draw OT and on it, beginning at 0, lay off the units 36 GRAPHS OF STATISTICAL DATA 37 of time to any convenient scale, say \" to 1 hr. At draw OS perpendicular to OT and lay off the units of distance to a scale of, say, \" to 1 mi. The point represents the starting point, mi. and hr. The point 1 of the path is represented at li on a A, 1 unit to right of OS and \ unit above OT. The point 2 of the path is represented at 2i on bB, 2 units to the right of OS and ^ + 1| = 2 units above OT. In a similar manner the remaining points of the path are represented on the diagram. s mi. f L ABCDEFGH j j / 10l ^ / ii / / f ^ i. / 1 X I 1 h / / - S O abcdefghtij FIG. 7. FIG. 8. 10i are now to be connected by The points 0, li, 2i, . . straight line segments. The broken line Oli, 2i, . . . 10i in a sense represents the travel of the man during a period of 10 hrs. To continue the study of the diagram we need to use certain terms which we shall now define. 28. The line OT is a base line and is called the axis of abscissas. In this particular case it is a time axis. The line OS is a meridian line and is called the axis of ordinates. In this 38 GRAPHIC REPRESENTATION particular case it is a distance axis. The distance of a point from the meridian is called its abscissa. The abscissa must be measured to the same scale as that to which 'the base line is laid off. The distance of a point from the base line is its ordinate. The ordinate must be measured to the same scale as that to which the meridian is laid off. The abscissa and ordinate of a point considered together are called its coordinates. Evidently the position of a point is determined if its coordinates are known. The difference of the ordinates of two points divided by the difference of their abscissas, taken in the same order, is called the slope of the straight line joining the two points. The slope is also called the pitch of the line. Resuming the study of the example answer the following questions: (a) What does the ordinate of any of the points li, 2i, . . . , of the broken line represent with reference to the travel of the man? (6) What does the abscissa represent? (c) What does the slope of any part of the line represent ? (d) Is the speed of the man constant for the whole time ? (e) What is the average speed of the man for the whole time ? Calculate the average speed for the first 3 hrs. For the second 3 hrs. For the last 3 hrs. 2. A rod 8" long is set upright on a level table in sunshine. The lengths of the shadow were measured at intervals as follows : Time of Length of L observation = t shadow = I j f 10 : 40, A.M. 7.79" 8- *i ^ 11:15, " 7.35" ' 11:45, " 7.25" 4- 12 : 00, M. 7.22" 2- 12 : 10, P.M. 7.22* 1- 1 1 1 1 , T 1 : 00, " 7.30" s o o 'II < o < g g 2:15, " 7.75" o IH i-l C IH 1 ^ fc' iH IM CO 2:30, " 9.81" FIG. ( J GRAPHS OF STATISTICAL DATA 39 Construct the graph using time as the abscissa, but draw a smooth curve instead of broken line.* Take time scale \" to one hour, starting at 10:00 A.M. Take length scale \" to 1". Draw a second curve with same time scale but with length scale \" to 1". What difference do you notice? Measure the ordinates at 11:00 A.M., and at 1:30 P.M., and determine the probable length of the shadow at these times. The graphic representation enables us to determine, approxi- mately at least, more values than are given in the measured data. The determination of new values between the old ones is called interpolation. Interpolation is a very useful applica- tion of the graphic representation. 3. The population of the U. S. by decades was as follows, in millions: 1790 3.9 millions 1830 12. 8 millions 1870 38. 5 millions 1800 5.3 " 1840 17.0 it 1880 50.0 M 1810 7.2 " 1850 23.2 it 1890 62.5 " 1820 9.7 " 1860 31.3 it 1900 76.3 1C 1910 92.0 (a) Construct a curve showing the probable population at any time, using tune as abscissa. (6) Connect the decade points of the curve by straight lines and determine the slope of each segment. What information does the slope give? (c) Construct a curve on same abscissas as (a) using the slopes found in (6) as ordinates. What information does this curve picture? 4. The following data were taken on a certain machine while in operation. The relative velocity of two moving parts was measured in ft. per sec. (ft./sec.), and the coefficient of friction for each velocity was measured. v = speed, ft./sec. = 135 7 10 15 20 w = coef. of friction = 0.15 0.122 0.104 0.092 0.079 0.066 0.058 * The curve is an attempt to represent the conditions between observed points, the broken line is not, except in special cases. 40 GRAPHIC REPRESENTATION Note. Coefficient of friction is the resistance to sliding of one part on another divided by the total pressure between the parts. Construct a friction curve with v as abscissa. What information about speed and friction does the curve reveal? 5. It costs $6 to make an article. The proprietor finds that he can sell in a given time the following numbers of the article at the corresponding prices: Selling price = S = $6 $12 $18 $24 $30 Number sold = N = 38,000 35,000 24,000 8000 6000 (a) Construct a sales curve using S as abscissa. (6) Calculate the profit at each price, P = N (S 6) and construct curve with same abscissas as in (a) with P as ordi- nates. (c) From the sales curve estimate the number that could be sold at $15 and at $20. Insert the values of P corresponding to these in the profit curve. Now draw the profit curve so as to pass through these points. (d) From the profit curve determine the selling price that will give the greatest total profit. (e) What business condition is indicated by the drop in the sales curve from $12 to $24? What inference can you draw from the nearly equal sales at $24 and $30? From those at $12 and $6? 6. The magnifying power (in linear dimensions) of a certain spyglass at different distances from the object viewed was measured as follows: Magnifying power = m = 9 8.3 8.1 7.6 7.5 7.6 7.6 Distance, meters =z=23 4 5 6 7 8 Construct the magnification curve, using x as abscissa. What general information does the curve reveal? What do you suspect regarding the correctness of the measurement 7.5? The curve should not be made to pass through this value as it is obviously wrong. 7. The sun's declination as given in the nautical almanac for 1915, near the time of the vernal equinox, is as follows: GRAPHS OF STATISTICAL DATA 41 March 18 noon South 1 05' " 19 " " 041' " 20 " " 018' " 21 " " 006' " 22 " North 30' " 23 " " 051' Construct the declination curve, using time as abscissa. Take a scale of T V" to 1 hr. This will enable the determination of hours with fair certainty. Determine the day and hour at which the curve crosses the axis of abscissas. This is the time of the vernal equinox. What is the time at which the sun is exactly over the equator? Note. The declination of a heavenly body is its distance north or south of the equator of the sky just as latitude is the distance of a place on the earth north or south of the equator. Declination and latitude are both measured in angular units, that is, in degrees. 8. If s is the amount (grams) of potassium bromide that will dissolve in 100 grams of water at t centigrade, construct a solubility curve from the following data, using t as abscissa. s = 53.4 64.6 74.6 87.7 93.5 t = 20 40 60 80 From the curve determine the probable amount that will dissolve at 10, 30, 50, 70. 9. With the same notation as in (8) the following data were taken for sodium nitrate: s = 68.8 72.9 87.5 102 t = -6 20 40 Construct the solubility curve and determine the probable amount that will dissolve at 10, 30. 10. Construct the temperature curve from the maximum daily temperatures at a certain city for the month of February. For convenience assume the maximum occurred at the same hour each day. Use time as abscissa. 42 GRAPHIC REPRESENTATION Day. Temp. Day. Temp. Day. Temp. Day. Temp. 1 28 8 -2 15 10 22 26 2 36 9 14 16 13 23 13 3 31 10 4 17 21 24 20 4 14 11 1 18 24 25 26 5 24 12 11 19 11 26 35 6 28 13 15 20 20 27 42 7 18 14 14 21 30 28 42 11. A 300-lb. projectile is shot from a 10" rifle with a 70-lb. charge of powder. The speed and powder pressure at different distances from the starting point were measured as follows : Distance, ft. Measured speed, ft./sec. Theoretical speed, ft./sec. Pressure, Ibs./sq. in. 0.41 436.0 450.8 34,752 0.50 501.7 505.0 34,577 0.82 657.0 661.0 32,000 1.06 750.0 748.2 29,648 1.65 911.0 908.0 24,400 2.06 992.8 992.5 21,403 2.26 1027.0 1027.0 19,302 2.74 1097.0 1098.0 17,580 3.06 1137.0 1139.0 16,121 7.06 1412.0 1420.0 7,022 8.26 1464.0 1464.0 5,476 10.15 1527.0 1516.0 4,269 Construct the speed curve and the pressure curve, using dis- tance as abscissa in both cases. In fact both eurves may be made on the same axes with different colored pencils. What information regarding speed and pressure at different points in the gun do these curves reveal? 12. Experiments with projectiles show that there is a varia- tion of muzzle velocity due to a variation of the temperature of the powder. The curve on the following page represents the functional relation of temperature of the powder and per cent change in muzzle velocity. For temperature above 70 the arithmetical correction of the velocity is added, for tem- perature below 70 it is subtracted. GRAPHS OF STATISTICAL DATA 43 Ti 44 GRAPHIC REPRESENTATION 1. The muzzle velocity of a certain gun is 980 ft./sec. when the temperature of the powder is 70. Show that 20 ft. should be subtracted from this velocity if the temperature of the powder is 41. 2. The muzzle velocity of a gun is 2250 ft./sec. when the temperature of the powder is 70. Find what correction should be made to this velocity when the temperature of the powder is 71.4. 3. The muzzle velocity of a gun is 2000 ft./sec. when the temperature of the powder is 70. Find what corrections should be made to this velocity when the temperature of the powder is 33.6. CHAPTER V RATIO, PROPORTION AND VARIATION 29. Whenever four numbers a, 6, c, d satisfy the equation (1) a/6 = c/d, these numbers are said to form a proportion or to be in propor- tion. The equation (1) contains two equal fractions or ratios. When we measure any quantity by comparing it with another of the same kind we are said to measure the first quantity by the second as a unit. The number which expresses how many units and parts of units result from the comparison is the measure of the quantity to the unit employed. 30. A ratio is defined as the measure of any quantity, ab- stract or concrete, when some other quantity of the same kind is the unit of measure. When the length of a room is measured with a foot rule (one-foot length) and there is obtained 24 as the measure or numerical value of the length of the room to that unit, it is said the length of the room is 24'. This implies ,^\ length of room _ _ . one foot length This is the idea involved in all measurements where definite units are available. If the length of the room is measured with a yard stick (one yard length), the measure of the room is 8. The room is said to be 8 yds. long. This implies ,* length of room _ _ one yd. length A ratio is to be regarded as a mere number without material denomination, that is, it is an abstract number. Since a ratio is essentially a quotient or an indicated division, it is con- 45 46 RATIO, PROPORTION AND VARIATION veniently represented by a fraction. A ratio is, therefore, subject to all the arithmetical operations ordinarily performed with fractions. 31. From (1) and some general axioms regarding equalities it is easy to establish the following fundamental and useful theorems: a c If T = -,, then it can be proved that: 1. ad = be, the product of the extremes equals the product of the means. 2. - = -j , proportion by alternation. C CL 3. - = - , proportion by inversion. Cl C 5. 6. 7. b a 6 c d i d proportion by division. proportion by composition and division. , a+c+e+0 a c e g b a + b. c d ' i d Tf a 6 b c c d~ e _ d' h +d+f+h ~b d f h 32. Variation. The statements, " x is proportional to y," "x varies as y," "x is directly proportional to y," "x varies directly as y," are all different ways of saying the same thing. This relation between x and y may be put in the form (1) . x = ky, * where A; is a fixed value, and x, y take any number of different values which satisfy this equation. The statements, ( 'x varies inversely as y," "x is inversely proportional to y, " are each equivalent to the equation k (2) xy = k or x = -, y where k is constant as in equation (1). J VARIATION 47 The statements, "x is jointly proportional to y and z" "x varies jointly as y and z," are each equivalent to the equation (3) x = kyz. From type forms (1), (2) there follows a single equation (4) ,-* Equations (1), (2), (3), (4) are to be regarded as special and convenient working forms of proportion. Many laws of nature are expressed in one or another of these forms. Illustrations. / = k , the law of gravitation; A = xy, the area of a rectangle; E = kn 2 , the law of mob energy. 1. Suppose it is known that s varies as t, and that s = 25, when t = 5. To find the equation of relation between s and t and to find the value of s when t = 10. Write by (1), s = kt. Substituting values given in the problem, 25 = k - 5, therefore k = 5. The desired equation is s = 5t. The value of s when t = 10 is s = 5 . 10 = 50. 2. The carrying capacity of pipes varies as the squares of their diameters (friction neglected). How many 6" pipes will carry as much as one 24" pipe ? Call the carrying capacity of a pipe C f and write (1) C = kd* for any and all pipes. For 6" pipes then, (2) Ce=? fc-6 2 48 RATIO, PROPORTION AND VARIATION (3) Also C 24 = k - 24? = 576 k. By equations (2) and (3) , Cu 576 (4) -^j- = -^- = 16, the required number of 6" pipes. 0$ 06 Note. In this case we did not determine the value of k. Its presence in the equations gave us all the advantage of knowing its value without finding it. 3. The weight of a sphere of given material varies as the cube of its radius. If a sphere of 1' radius weighs 600 Ibs., what will a sphere of radius 5^' of same material weigh? Note. Solve this and the following by the methods of Ex. (1), (2), above. 4. The capacity of a cylindrical tank varies as its height when the diameter is fixed and as the square of the diameter when the height is fixed. A tank 1' in diameter and 1' high has a capacity of 5.83 gallons. Find the capacity of a tank 8' in diameter and 30' high. Note. Write C = k> &- h. 5. The value of a diamond varies as the square of its weight. A diamond worth $600 is cut in two pieces whose weights are as 1 to 3. What is the value of each piece? 6. The illumination from a light varies inversely as the square of the distance from the light. If an object 10" from the light be moved 10 (Vo 1)" farther away, what is the ratio of the final to the original illumination. (Assume I as the original illumination.) 7. The cross section of a chimney should vary as the quan- tity of fuel used per hr. and inversely as the square root of the height. The cross section of a chimney 150' high is 30 sq. ft., the quantity of fuel used per hr. is 15,000 Ibs. Find the cross section of a chimney 100' high connected to a furnace using 5000 Ibs. of fuel per hr. 8. A solid spherical mass of clay 4" in diameter is moulded into a spherical shell whose outside diameter is 6". What is SUPPLEMENTARY EXERCISES 49 the inside diameter of the shell? It is given that the volume of a sphere varies as the cube of its diameter. 9. A safe load on a horizontal beam supported at its ends varies as the breadth and as the square of the depth and in- versely as the length of the beam. A beam 2" x 6" x 12', on edge, will sustain a load of 700 Ibs. What load will a beam of the same material 3" x 9" x 18', on edge, sustain? 10. The weight of a body above the earth's surface varies inversely as the square of its distance from the center of the earth. If a body weighs 150 Ibs. just outside the surface, how high must it be raised so its weight will be 30 Ibs., the radius of the earth being assumed to be 4000 mi? 11. A tree casts a shadow 70' long on level ground. At the same time a 10' pole casts a shadow 9' long. Find the height of the tree. 12. The velocity of a falling body starting from rest varies as the time. At the end of 2 sec. the velocity is 64.32' per sec. What is the formula holding between velocity and time and what is the velocity at the end of 4 seconds? Note. Assume v = k t. 13. The interest on a given principle varies jointly as the time and rate. $500 yields $25 in two years. What is the rate ? 14. If the amount of fuel required to heat a house varies as the square of the difference in temperature in the house and outside and if when the thermometer reads C outside, and the temperature inside is 20 C, it requires 1000 cu. ft. of gas per hr. to heat a certain house, how much gas would be neces- sary to heat the house to the same temperature when the ther- mometer outside dropped to 10 C ? SUPPLEMENTARY EXERCISES 1. If y varies as x and if x = 4 when y = f , find y when x = 3. 2. If y is proportional to x and if x = when y = f , find x when y = 7. 3. If y varies directly as x and inversely as z and if y = 8 when x = 12 and 2 = f , find z when y = 2 and x = 6. 4. If y varies jointly as x and z and if y = f when x = 5 and z = f , find z when y 3 and x = 6. 50 RATIO, PROPORTION AND VARIATION 5. If y varies as x and x = 6 when y = 54, what is the value of y when x = 8? 6. If x varies directly as y and inversely as z, and x = f when y = 20 and z = 10, what is the value of x when y 9 and 2 = 20 ? 7. If x 4 is directly proportional to y 3 and x = 4 when y 4, what is the value of x when y = 9 ? 8. If 2 x 3 is proportional to y + 6 and x = 4 when y = 3, what is the value of y when x = 5? 9. The hypotenuse of a right triangle is 100 ft. long. Find the other sides, if their ratio is 3 to 4. 10. The stretch in the spring of an ordinary spring balance is propor- tional to the weight (force) applied. If a force of 4 Ibs. stretches it one inch, how much will a force of 17 Ibs. stretch it? 11. How would you graduate (divide) a scale for the spring balance in the above problem? 12. The area of a circle varies as the square of its diameter. How is the diameter affected when the area is doubled? How is the area affected when the diameter is doubled? Give work showing reasons for your answers. 13. If two pulleys are connected with a belt prove: (a) That the number of revolutions that they make per minute are to each other inversely as their diameters. (6) That their radii are to each other inversely as their speeds. 14. If x varies as y, then x = ky and conversely, if x = ky, then x varies as y. Show that the speed of a point on the rim of a pulley varies as its diameter. 15. If in problem 13 the driving pulley is making 20 revolutions per min- ute, and its diameter is 10 inches, find the number of revolutions per minute of the second pulley if it is 4 inches in diameter. 16. According to Boyle's law of gases, pressure (p) times volume (v) is constant. How does the pressure vary with the volume? Show graphi- cally the relation between (p) and (v) if v = 1 cu. ft. when p = 25 Ibs. per sq. in. 17. Given that: L = 2-xrh and L' = A=2irr(r + h] and A' = 2^' (r' + fc'), formulas which represent the total area and lateral area of two right circular cylinders. Show that IL - A = z! = ^1 L' A' r'* h 1 *' 18. In reading contour maps the question arises whether a station B is SUPPLEMENTARY EXERCISES 51 visible from some station A. The problem is to determine whether an intervening height of ground C obstructs the line of sight from A to B. FIG. 10. The horizontal distance between A and B as scaled on the map is 1500 yds., between A and C is 900 yds., and height of B above datum plane (horizontal plane through station A) is 80 ft., height of obstacle C above datum plane is 50 ft. Determine whether A is visible from B. By means of the figure and theorem 27, Chapter II, it is seen that: and solving 900 : 1500 = h : 80 h = 48. Therefore, the line joining stations A and B would pass station C at a height of 48 ft. above the datum plane. Since obstacle C is 50 ft. above this plane it obstructs the vision from B to A. 19. Check the above problem by drawing the figure to scale. 20. In problem 18 substitute y for 900, x for 1500, H for 80, and solve the equation for h. Draw a figure and indicate the distances x, y, H and h. 21. Using the result of problem 20, find whether C would obstruct the line of sight from B to A if x = 2100 yds., y = 1400 yds., H = 130 ft. and the height of C above datum plane is 95 ft. CHAPTER VI THE RECTANGULAR COORDINATE SYSTEM: GRAPHS OF EQUATIONS; FORMULAS 33. Two lines are selected, intersecting at right angles, in the plane of the paper (they are usually chosen the one hori- zontal and the other vertical). The position of a point is known if its distances from these two lines are known. Let FIG. 11. XX' and YY' intersect at right angles in 0. The position of P is known if Om = x and mP = y are known. The point P will often be designated by P (x, y) or by (x, y), at pleasure. It is noted that the abscissa x is always written before the ordinate y and this arrangement holds no matter what letters are used to represent the coordinates. 52 GRAPHS OF FUNCTIONS AND EQUATIONS 53 Since the point P may lie either to the right or left of YY' and above or below XX', we must have further conventions. The following have been universally adopted: The ordinates of points above the XX' axis are positive. The ordinates of points below the XX' axis are negative. The abscissas of points to the right of YY' axis are positive. The abscissas of points to the left of YY' axis are negative. The point is called the origin of coordinates or, for short, the origin. Both its coordinates are zero. 1. Locate the following points in rectangular coordinates: (2, -5); (-4, 1); (-3, -3); (4,5); (1,8). 2. What is the abscissa of any point on YY't What is the ordinate of any point on XX' ? 3. What is the ordinate of any point of a line parallel to XX' and 6 units above it ? - On what line do all points whose abscissas equal 3 lie? 34. Graphs of functions * and equations. To construct the graph of any function, say 3 z 2 4 x + 5, write it equal to some symbol, say y, obtaining an equation, (1) 2/ = 3z 2 -4z + 5. Select one of the symbols x and y for the abscissa and the other as the ordinate of points on the graph. In this and similar cases, x is usually the abscissa and y the ordinate. Every equation like (1) expresses a relation between different number pairs. Every number pair are the coordinates of a point. One number of each pair may be chosen at pleasure, the other number of the pair must be calculated from the equation. Any number of such number pairs may thus be determined. The corresponding points can be located on the diagram. Having located several points, draw a smooth curve through these points as in Chapter IV. This curve is called the graph or the locus of the equation. It must be remembered that the coordinates of all points on * For definition of function see Chapter VII. The term formula might be used at present. 54 the graph must satisfy the equation. On the other hand every number pair which satisfy the equation must be coordinates of a point on the graph. To be sure, the absolutely exact graph cannot be drawn without determining the coordinates of every point on it. This, obviously, would be impossible for it would involve endless calculation to obtain even a small portion of the graph. But by carefully determining a few well-selected points of the graph and then drawing a smooth curve through these points, it will be found that a very close approximation to the true graph is obtained. This approximate graph is sufficiently accurate to permit interpolation and to furnish a good basis for a study of the equation it represents and of the true graph. The process of con- structing the graph of equation (1) is indicated below. Choose values of x and calculate the cor- responding values of y from the equation, x= -2, -1, 0, + 1, +2, +3 y = +25, +12, +5, +4, +9, +20 The annexed figure shows the approximate graph X which gives a good idea of the true graph. Owing to the large values of y, a smaller scale is used for ordinates than for abscissas in order to reduce the space necessary to draw a sufficient portion of the curve. This distortion modifies the form of the curve but does not affect its fundamental nature. It reduces the distance between the calculated points of the curve, see 17. FIG. 12. GRAPHS OF FUNCTIONS AND EQUATIONS 55 2. Construct the graph of the equation 2 x + 3 y = 5. Assign values to x and calculate values of y from the equation. x= -3 2 5 8 y = gl 1 ^2 JJ2 This locus is a straight line. What is its slope? At what value of x does the line cross the axis XX"! This value is the x- intercept of the line. At what value of y does the line cross the axis YY'f This value is the ^-intercept of the line. Y FIG. 13. A straight line is always associated with an equation of the first degree in two unknowns. 3. Construct the graph of the equation x 2 + y 2 = 25. As in the preceding cases choose values of x and calculate the corresponding values of y from the equation. x= -6 -5 -4 -3 -2 -1 y= (i)* 3 4 4.6 4.9 x = +1 +2 +3 +4 +5 +6 y=4.9 4.6 4 3 (i)* It is noted that to every value of x there correspond two equal but oppositely signed values of y. The curve is, therefore, * (i) represents an imaginary quantity. 5 56 THE RECTANGULAR COORDINATE SYSTEM FIG. 14. symmetrical about the axis XX'. If values of y are chosen and values of x calculated from the equation it will be found that to every value assigned to y there cor- respond two equal and opposite values of x. Therefore the curve is symmet- rical about the axis YY'. Conse- quently the curve is symmetrical about the origin. This curve is a circle. Its radius is 5 and its center is at 0. A circle is always associated with an equation of the form x 2 + y z = a 2 , that is, the sum of two squares equals a constant. What is the nature of the values of y for all values of x < 5 ? y " x > 5? x y < -5? " " " x " y > 5? It is seen that no real values of x or y outside the limits 5 and +5 will satisfy the equation. This means that no points of the curve can lie outside these limits. 4. The corresponding values of two symbols x and y are such that the square of one equals the square of the other increased by 1. Write the equation and construct the graph. Save graph for Ex. 7. 5. The corresponding values of two symbols are such that the sum of their squares equals 49. Write the equation and construct the graph. Save graph for Ex. 7. 6. Construct the graphs of y = 2 x and of y = 2 x + 6 on the same axes and to same scale. What likeness and what difference do you notice regarding the lines? What likeness and what difference regarding the equations? What is the slope of each line ? What is the coefficient of x in each equation? Can you tell the slope of a straight line from its equation without drawing the line? How? 7. Determine intercepts on the axes of all the lines in the last three exercises. EMPIRICAL EQUATIONS 57 8. Construct the graph of x + 4 y = 14 and determine the slope and the intercept on the axes. 9o Construct the graphs of rc/3 + y/5 = 9 and 8 x *- 4 y + 4 = on the same diagram. Determine by measurement the coordinates of the point of intersection of these graphs. Solve the given equations as simultaneous equations for x and y. Compare the results with the coordinates of the point of inter- section. Explain. 10. Treat the equations x 2 + y z = 25 and x y = 1, in a manner similar to that directed in Ex. 9. 11. Treat the equations x + y = 6 and x y = 1 in a manner similar to that directed in Ex. 9. 12. Draw the triangle whose sides lie in the lines represented by the equations: x + y = 4; 2x + y = 2; x y = 6. Note. The vertices of the triangle will be the three points of intersection of the lines in pairs. 13. A sphere of wood 1' in diameter sinks in water to a depth determined by one root of the equation 2 x 3 3 x 2 + 0.657 = 0. Note. The desired root will be one of the x-intercepts of the curve, i/ = 2z 3 -3z 2 + 0.657. Explain. 35. An important use of graphs is found in the determina- tion of empirical formulas expressing laws of nature. From observations made in the laboratory or in the field several number pairs are measured. From these a curve is constructed. This curve represents to the eye the relation between the numbers of each of the number pairs. The form of the curve may often suggest to the experienced mathematician the general form of an equation of which the curve is the graph. It remains to determine the constants of this equation. Some- times only an approximately correct formula can be determined at first. This formula is subject to later correction by addi- tional observations and by the application of least squares. A few examples will illustrate the meihod of procedure. 1. Let us attempt to find an equation for problem 6, 28, Chapter IV. The form of the curve suggests to one expe- rienced in the art, an equation of the form xy ~ k or y = 58 THE RECTANGULAR COORDINATE SYSTEM - + 6 as possibilities where either k or else a and b are to be determined. Let us try the second form and write (1) m = l + b. There being two unknown constants we shall need two equa- tions. These are obtained by substituting in (1) two pairs of observed values of m and x. Thus (2) 9 = ! + & > (3) 7.6 = ^ + 6. o Solving these equations simultaneously for a and 6 gives a = 4.7 and 6 = 6.7. Substituting these in equation (1) gives as the desired formula: (4) m = ^ + 6.7. To check the validity of this formula for values not included in determining a and b, put x = 4 and m turns out to be 7.9. The corresponding observed value is 8.1. The formula gives fairly good results considering the nature of the quantities concerned. 2. An innkeeper finds that if he has G guests per day his expenses are $E and his receipts $R. His books furnish the following data: G = 210 270 320 360 #=83 97 108 117 R= 79 106 132 149 Using G as abscissa construct two graphs, one for E and one for R on same axes and same scale. These lines appear to be straight lines. This fact suggests an equation for each of the form y = mx + 6. Write EMPIRICAL EQUATIONS 59 (1) (2) E = R = + bi. + 62. The values of m\, mz, &i, 62 may be determined easily from the graph. The slopes are mi, d'c' dc Measure m i = ji = 0.203; mz = jTr, = 0.444. The ^/-intercepts are ob = bi = 32; ob' = bz = 35. These values are to be regarded as close approxi- 150- -Id'. FIG. 15. mations, only. The equations (1) and (2) now become, by substituting these values, (!') E = 0.203 G + 32. (2') R = 0.444 G- 35. (a) What number of guests is just sufficient to pay expenses ? (6) What are the expenses when G is zero? What are the profits when G = 360? How can profits and losses be deter- mined from the graph? 60 THE RECTANGULAR COORDINATE SYSTEM Equations (!') and (2') can be obtained algebraically. Sub- stitute two pairs of values of E and G in (1) and solve for mi, 61. Then substitute two pairs of values of R and G in (2) and solve for mz and 62. Thus from (1) 83 = Wi 0.210 + 61, 117 = mi- 360 + 6!. Whence mi = 0.233 and 61 = 35. In a similar manner with Eq. (2): 79 = ma 210 + 62, 149 = mz 360 + 62. Whence mz = 0.466 and 6 2 = 29. These values are in fair agreement with the ones determined above. 3. The latent heat of steam at 6 C. is L. Construct a curve from the values given below and determine an equation of the form L = md + b. 4. The height h, above the earth's surface and the corre- sponding barometer reading p, are as follows: h (ft.) = 886 2703 4763 6942 p (in.) = 30 29 27 25 23 Construct the graph and determine an equation of the form p = Ae kh . Note. Take the logarithm of both sides of the proposed equation and proceed as in previous cases. Thus log p = log A + kh log e, e = 2.718. Determine first, log A and k as in previous cases. The value of A and k are then to be substituted in the proposed equation p = A& h . 5. Determine the equation of the form y = ax 2 + bx + c of the curve which passes through the points given by x= 2 3 4 5 y = 10 -6 -5 10 EMPIRICAL EQUATIONS 61 6. Construct the curve and determine the equation of the fox-m y = y? + bx* -+- ex + d from the following data: x= -2 -1 1 2 3 y = -8 -1 1 2 27 7. Construct the graph and determine an equation of the form y z = ax + 6 from x = 2 4 y = 4 10 Determine whether or not the points (0, 0), (3, 7) are off the curve. 8. Determine an equation and draw the curve from x = Q 136 t/ = 6 5.9 5.3 The equation is to be of the form ax 2 + by 2 = c 2 . 9. For an ideal gas Boyle's law says that the product of the pressure and volume of a given mass of gas at constant tem- perature is constant. Form an equation from this statement and draw the graph. Determine an equation from the follow- ing: Pressure, inches of mercury = 130 45 60 75 90 105 Volume, cubic centimeter = 100 66.6 50 40 33.3 28.5 10. The intensity of illumination from a light varies inversely as the square of the distance from the light. Write this in the form of an equation and draw the graph. Determine an equation from L = 100 50 25 5 Z>= 10 14.14 20 44 Note. If desired the subject of empirical formulas may be continued at this time by taking up the work, which appears in a later chapter, on the applications of logarithmic and semi- logarithmic paper in determining certain types of empirical formulas. CHAPTER VII NUMBERS, VARIABLES, FUNCTIONS, LIMITS 36. Numbers. (a) The numbers 1, 2, 3, ... used in ordinary counting are called the natural or absolute integers. The idea of positive and negative does not belong to them. They answer the question, "How many?" We associate with the natural integers all fractions formed with them, such as, |, f, -VS .... The natural integers and the associated fractions constitute the number system of ordinary arithmetic. They are sometimes called non-directed numbers. They are some- times, but with questionable propriety, identified with the positive numbers of algebra. (6) The idea of positive and negative numbers is a relatively modern notion. Certain natural phenomena, scientific meas- urements and the fact that subtraction was not always possible with the natural numbers suggested the idea of "opposite" numbers or positive and negative numbers. (c) A rational number is one that can be expressed as the quotient of two integers, positive or negative. Some rational numbers are expressible in decimal form as f,"| = 0.4, | = 0.125, etc. Some rational numbers cannot be exactly ex- pressed in decimal form, as f = 0.666 . . . , | = 0.1111 .... (d) Irrational numbers jesult from certain operations on rational numbers. Thus \/2 = 1.4142 ... is irrational. For it cannot be expressed as the quotient of two' integers. The numbers V3, v^, TT = 3.14159 . . . are examples of irrational numbers. The logarithms of most numbers are irrational. (e) All the numbers so far mentioned come under a more general class called real numbers. The name is derived from their association with ordinary affairs and by contrast with a class of numbers defined below. 62 A VARIABLE 63 (/) An imaginary number is one that arises in the attempt to take an even root of a negative real number. Thus \/ 2, v' 3, . . . are imaginary numbers. Remark. It should be noted that the name "imaginary' applied to numbers reflects an attitude of mind, existing for- merly, toward such numbers. Recent developments have shown that term is unfortunate. For imaginary numbers have come to have a very real meaning in scientific investigations. Attention seems first to have been directed to imaginary numbers in the study of quadratic equations. (gr) The sum of a real number and an imaginary number is called a complex number. All complex numbers are of the form a 6 V^I where a, 6 are real. Thus 2 + V^9 = 2 + 3 V^; -| + | V^3 = -| + V3 V^l are complex numbers. Use will be made of complex numbers in a later chapter. It is to be kept in mind that in practical calculations involving irrational numbers it is necessary to use a rational number nearly equal to the corresponding irrational number. Thus for \/2 the rational number 1.4142, which is correct to five figures, may be used, and similarly in other cases. 37. (a) A variable is a symbol to which, in a given problem, may be assigned an indefinitely great number of values, and which may be employed in calculations as a number. The values assigned to a variable may be assigned in accordance with some law of nature or in accordance with some arbitrary mode of thought. May x and y have more than one value each in the equation y 2 x = 4? Does 2 ever have any value different from 2? Are x and y variables or fixed in the equation ? Is 2 variable or fixed? Can you define a constant? (6) When a variable assumes or may assume every assignable value between two given constant values the variable is said to be continuous or to vary in a continuous manner, in the inter- val between the given constants. 64 NUMBERS, VARIABLES, FUNCTIONS, LIMITS The totality of values between two given values is called an interval. If the given values are included the interval is closed, if not the interval is open. If a variable may not assume all values in a given interval it is not continuous in that interval, though it may be continuous in' parts of the interval. Suppose a bottle of ink stands, uncorked, on the table for several days or weeks, and suppose it is not disturbed hi any way. Is the amount of ink in the bottle from time to time the same? Is the quantity of ink constant? Does it vary con- tinuously ? Suppose wheat is $1.75 per bu. today, $1.90 yesterday and $2.00 tomorrow. Is the price of wheat constant? Does it vary continuously? (c) Any set of numbers taken in some definite order is called a sequence. If there are infinitely many numbers in the set the sequence is called an infinite sequence. If there are only a finite number of numbers in the set it is a finite sequence. Thus 1, 3, 4, 7, 9, 15, form a finite sequence. But 1, 1.1, 1.11, 1.111, . . . form an infinite sequence. A sequence may increase or decrease. That is, the successive numbers may be larger than or smaller than the preceding, respectively. (d) Let x be a variable and a some constant. Then if a x assumes its sequence of values in order, there comes a stage, such that, for all subsequent values of x, the numerical value of a x becomes and remains less than any assigned small value e, then a is called the limit of x. This definition is expressed in symbols as a = limit x, or x > a or x = a. All these forms may- be read a is the limit of a; or a; approaches a as a limit. A variable may or may not become equal to its limit, depend- ing on the nature of the law of its change. A VARIABLE 65 The difference between a variable and its limit is a variable whose limit is zero. A variable whose limit is zero is called an infinitesimal. A variable that may increase without limit is said to have no limit or with less propriety to have infinity for its limit. (e) Let y be a variable and x another variable. If to every value assigned to x there corresponds a definite value assumed by y, then y is a function of x. More concretely but less exactly it may be said that the values of y, the function, depend upon the values of x, the vari- able. From this notion we have the terms dependent variable or function and independent variable or merely variable. In the above definition x is the independent variable and y the dependent variable. In dealing with equations and their graphs it is customary to regard the abscissa as the independent variable and the ordinate, the dependent variable or function. 38. When it is desired to express briefly the fact that y is a function of x we may write: y = f(%) (read y equals/ of x or /function of x) or y = 4>(x), etc. This is a notation used for convenience. Suppose y is defined as a function of x by the expression, y=f(x) = 4x* -Qx + 4* This equation defines f(x) for this particular case and during the discussion of this function f(x) is understood as a brief way of writing the polynomial 4 x 3 6 x + 4. When particular values are substituted for x the fact is indicated by /(3) = 4 3 3 - 6.3 + 4, . /(-l)=4.(-l)'-6.(-l) + 4, /(a) = 4 a 3 - 6 a + 4, etc. If f(x) is any function of x and if /(a) /(z) * 0, when x > a, f(x) is called a continuous function of x at the value x - a. It is assumed that x > a, either by decreasing or by increasing values. This is expressed by writing | /(a) f(x) \ * This equation is called a functional equation between x and y. 66 NUMBERS, VARIABLES, FUNCTIONS, LIMITS where the "| |" indicate the absolute value; that is, the numerical value without a plus or minus sign. 39. It is one of the chief problems of mathematics to dis- cover and to study functions of variables. This study is most easily carried on when the function is expressed as an equation between the function and the independent variable. Such an equation is called a functional equation. By studying the equation the mathematician learns the character or properties of the function which the equation expresses. When any natural phenomenon becomes so well known that it can be described by a functional equation, it becomes a part of mathematics. The mathematician may discover further facts and peculiarities of the phenomenon. In this way science and its applications have been greatly extended. 40. The difference between two successive values of a variable is called an increment of the variable. Let x\, Xz be two successive values of the variable x. Then Xz x^ is the increment of x. The symbol Arc * is used to represent the increment of x. Thus Xz Xi = Ax. When Xz > x i} Ax > 0. When Xz < x i} Ax < 0. Similar definitions and notations apply to any variable and to functions. 41. Special forms and limits. Theorems. a I. Lim II. Lim III. Lim IV. Lim = 0, where a is a definite number. = oo , where a is a definite number. = 0, where a is a definite number not zero. = oo , where a is a definite number* not zero. X The student can easily satisfy himself concerning the rea- sonableness of these theorems by arithmetical methods. For example, consider the sequence of fractions with the same numerators but with increasing denominators, * Read delta x. SPECIAL FORMS AND LIMITS 67 From our knowledge of division in arithmetic it is evident that the successive fractions are smaller and smaller in value as we proceed with the sequence. That is, they are nearer and nearer zero. They are approaching zero. Below are given the usual proofs of the above theorems. 42. I. The limit of - as x becomes infinite is zero, that is, lim x Suppose = 0, where a is a definite number. < e or \-\ < e \x\ where e is an arbitrarily small number, not 0. Then \a\<e\x\ and |*|>^ Therefore, for any assigned value of e, e ?* 0, we can calculate a value of x such that increasing values of x, < . It follows that for indefinitely satisfies the definition of limit of a variable and has zero for its limit. II. Lim = oo, where a is a definite number. It will be sufficient to show that if a; is chosen sufficiently large the inequality > M can be satisfied, however large M is chosen. For multiplying both sides by | a| gives \x\ >\a\M. It is only necessary then to choose x > \ a \ M in order to ensure the inequality III. Lim > M , however large M may be. = 0, where a is a definite number not zero. It will be sufficient to show that if a; is chosen sufficiently 68 NUMBERS, VARIABLES, FUNCTIONS, LIMITS \x\ small the inequality, r -4 < e, however small e be chosen, will be I a l satisfied. Multiplying both sides by | a | gives I a; | < e |a|. Now if a and e are given x is determined at once so as to satisfy \x\ the condition ][<. a IV. Lim = oo, where a is a definite number not zero. x->0 It will be sufficient to show that if x be chosen sufficiently smaD the inequality t- 4 > M , however large M is chosen, holds. \x\ Multiplying both sides by | x \ gives \a\>M\x\. Dividing now by M, This value of x will ensure the first inequality. 43. It is often desirable to know the limit approached by an expression when the variable approaches a given value. The preceding theorems are useful for this purpose. x 2 1. What is the limiting value of -5 : -.. when x >2. x 2 4x + 4' By direct substitution of x = 2 in the function the result is 0/0. This result can have no meaning. But it is noticed that this expression is not in its lowest terms. For x-2 1 s 2 -4z + 4 x-2 Now as x = 2, the result is oo , by IV. The expression 0/0 may be assigned any value. For, write x j - = k y NUMBER PAIRS 69 and let x and y > in such a way that this equation always holds. Multiplying by y gives x = ky. This equation is satisfied even when x = and y = 0. But k was arbitrary, therefore may have any value and is indeter- minate. 2. What is the limit of ^ 5 ^ as x oo. By direct substitution the result is ^/^ which is equally as indeter- minate 0/0. But if numerator and denominator be divided by y? the expression becomes, x x Now as x > oo ah 1 the terms of the numerator become 0, by I. The denominator becomes 1. Therefore the value of the fraction becomes 0, by III. x 1 3. What is the limit of J -^ , when x > 1 ? o^ l' 4. What is the limit of z 2 6 x + 2 when x 0, x 2, x > 10, respectively ? ^ 5. What is the limit of e x , when x > oo ? 6. What is the limit of x + -, when x > oo ? Whenz >0? x x 2 7. What is the limit of -^ -- r, when z > 0? X ^~ X 8. What is the limit of e~ x , when x > ? When x > oo ? 44. It should be noted that in much of the work of previous chapters we were concerned with number pairs. We selected one number of a pair and found the other by observation or calculation. This correspondence of number pairs satisfies the definition of function. Suppose (xi, t/i), (a*, 3/2), (x a , ya), . . . , (x n , y n ) be a set of number pairs, such that as a variable x 70 NUMBERS, VARIABLES, FUNCTIONS, LIMITS * assumes the values Xi, x^, . . . , x n , another variable y assumes the corresponding values, y\, y%, . . . , y n , the variable y is to be regarded as a function of the variable, x, over the given set of values. If each pair of values of x and y be regarded as the coordinates of a point, a curve drawn through these points represents more or less approximately the function in question. In fact the curve may be regarded as defining, approximately, a function, a few of whose values are known, viz., the points used in draw- ing the curve. By measuring the abscissas and ordinates of other points of the curve we may determine any number of other values of the variable and the corresponding values of the function, approximately. It is seen that the graphic representations of previous chapters represent or define, ap- proximately at least, functions, whether we know the functional equations or not. In some cases we were able to discover an equation from the graph. It is essential to progress in mathematics and its applications that we recognize the use of the graph as a means of studying functions and of discovering the corresponding functional equations, and that we learn to study any function by means of both its equation and its graph. In the next chapter we shall study a class of functions which have a wide range of application, not only in mathematics but in science and engineering as well. In later chapters we shall study still other kinds of functions. CHAPTER VIII THE TRIGONOMETRIC FUNCTIONS 45. Problem. It is desired to know the height of a tree, BC. It is found that the distance AC and the angle CAB can be measured. From this data it is desired to find the height of the tree, but these measurements alone will not be sufficient FIG. 20. for the purpose. The additional measurements AC' B'C' are, therefore, taken. This is done by means of the upright pole B'C' such that AB'B, the line of sight, is a straight line. Now from the similar triangles AC'B' and A CB the proportion B'C' BC or (2) AC' AC 1 B'C' BC = ~T7if ' AC can be written. Equation (2) shows that BC depends on A C and the ratio B'C' /AC'. This ratio evidently depends in some way on the angle CAB = 6. The frequent occurrence, in science and engineering, of situations similar to this caused 71 72 THE TRIGONOMETRIC FUNCTIONS mathematicians to search for the functional relation of the ratio to the corresponding angle. By careful measurements and calculations the values of the ratio and the corresponding angle have been tabulated for convenience in solving problems. The ratio B'C'/AC' is called the tangent ratio of the angle 0. Having a table of such ratios for different angles it is easy to calculate the height of the tree from the original measurements of AC and the angle 6 = CAB. Equation (2) may now be written (3) BC = AC (tangent of 6) or more briefly (4) BC FIG. 21. In this chapter we shall study the theory and use of the tangent ratio and other related ratios. The methods developed will furnish ways of solving problems of the highest importance in science and engineering. 46. An angle will now be regarded as generated by the rotation of a straight line about one of its points from some HESSLER'S DEFINITIONS 73 initial position to some final or terminal position. The angle will be included between lines lying in these two positions with its vertex at the point of rotation. For the purpose of formulat- ing the definitions and fundamental relations the line OX in Fig. 21 will be taken as the standard initial line or position. Let OP rotate from OX, about 0, counterclockwise. Let OP', OP", OP" 7 , OP IV be successive positions chosen as OP rotates. These lines are the terminal lines of the angles, a\, a 2 , a 3 , at, respectively, as shown in the figure. Counterclockwise rotation is regarded as positive and clock- wise as negative. This convention gives rise to positive and negative angles to correspond. The angle a& is negative. 47. Hessler's definitions of the trigonometric functions (ratios) are as follows: (See Fig. 22.) Name. Symbol. a, Quad. I. Quad 2 . II. Quad! III. Quad! IV. sine of a sin a cos a tana cot a sec a CSC a vers a covers c <+>! (+)- = 1 CO * = 1 sir ( ^ ( } h 2 X2 *^2 <-> 1 S 1 a .hi (+) (-)f-; 9t (-)- cosiile of a tangent of a cotangent of a secant of a cosecant of a versed sine of a coversed sine of a ... Note the signs of the ratios in the different quadrants. These depend on the signs of x and y in the various quadrants. The above table and figure must be memorized as a basis of future work. The haversine is defined as follows: hav x = | vers x = $ (1 cos x). By use of tables of natural haversines and their logarithms the solution of many of the problems in nautical astronomy is greatly simplified. 74 THE TRIGONOMETRIC FUNCTIONS An angle is said to be in or to lie in that quadrant in which its terminal line lies. Thus i is in the first quadrant, and 3 is in Quad. II the third quadrant. Functions of negative angles are defined exactly as for positive angles. Thus sin a 5 = yt/hi, etc. 1. In what quadrant is the angle 140? 210? 85? 240? 2. What is the sign of each function of each angle in Ex. (1) above ? 3. Draw an arc of 90 with a radius 10 cm. Divide the arc into 10 arcs. Measure the coordinates of the points of division. Divide each ordinate by the radius, 10 cm., and tabulate the quotients. Divide each abscissa by the radius and tabulate. Divide each ordinate by the corresponding abscissa. Arrange all these ratios in a table with the corresponding angles and compare them with the values given in the table of sines, cosines and tangents for the same angles, respectively, in the back of the book. FIG. 23. FUNDAMENTAL FORMULAS 75 48. From the definitions of 47 and the Pythagorean theorem the following fundamental formulas which hold in all the quad- rants are derived: (1) (x\ 2 s) + fy\ 2 x 2 (h) = h* ' +-r2 = = L ' cos 2 a + sin 2 a = 1.* nr hr (2) y X 1 x cot a y (3) h i sec a = or cos a sec a = 1. x X cos a h h 1 1 fA\ w y y sin a h ft(\ (h V i-' i 2 - x 2 _ y 2 _ 2 _ _ 2 ' (6) h y sin a (7) - = - . .'. - = tan a. x x cosa h The formulas in the right-hand column must be memorized. 1. Given sin A = 1/2, find cos A and tan A. Construct a right triangle whose hypotenuse is 2 and whose vertical leg is 1, as in Fig. 24. Calculate x = A/2 2 - I 2 = V = 1.732. 1 7*32 cos A = - - = 0.866, by the definition, 47. i sin A 0.5 _ ___ tan A = - T = jr^^ = 0.577. cos J. 0.866 2. Given sec A = 3, find cos A. and sin A. Construct a * The symbol sin 2 a is used for (sin a) 2 , being more convenient. notations are used for the other ratios. 76 THE TRIGONOMETRIC FUNCTIONS right triangle whose hypothenuse is 3 and horizontal leg is 1 as in Fig. 25. Calculate y = \/3 2 - I 2 = Vs = 2 \/2 = 2.828. y 2.828 A " *- O ^33 sin 4 y COS ^x == ~~" ~~ "~~ UOOO* bill \. y /I O /v O 3. Given sin A = f , calculate cos A, tan A, and sec A. 4. Given tan A = *, calculate sin A and cos A. 5. If one leg of a right triangle is 24 and the hypotenuse 30, find all the functions of the angle adjacent the given leg. 6. One leg of a right triangle is half the hypotenuse. Find all the func- tions of the angle ,*,_ = 0.9427. opposite the leg. FIG. 24. C=1 FIG. 25. 7. If sec = 1.5, find tan 6 and sin 6. 8. If tan 6 = 2.5, find sin 6 and cos 6. 9. The hypotenuse of a right triangle is 12, the base is 8, find the sine and tangent of the angle opposite the base. 10. If the tangent of A is 1, find the secant of A and the sine of A. In order that the student shall become sufficiently familiar with the seven fundamental formulas and their use the exercises below should be solved. The given equations are identities. Either or both sides are to be modified by various substitutions from the fundamental formulas so that both sides shall appear to be identically equal. 1. Cos A tan A = sin A. Write this in the form, using Eq. 7, i sin A cos A -r = sin A. cos A IDENTITIES 77 Reducing the left side this becomes sinA = sin A which is known to be identically true for all values of A. 2. Sec A sin A = tan A. (Use equations 3, 7.) 3. Cos A esc A = cot A. 4. (sin A cos AY = 1 2 sin A cos A. (Expand and use Eq. 1.) 5. sin A cot A = cos A. 6. esc A tan A = sec A. 7. cos A /(sin A cot 2 A) = tan A . 8. (sin 3 A cos 3 A) = (sin A cos A) (1 + sin A cos A). 9. (acosA +6sinA) 2 + (asinA -6 cos A) 2 = a 2 + b*. 10. sec 2 A + esc 2 A = tan 2 A + cot 2 A + 2. 11. cos A/(l tan A) + sin A/(l cot A) = sin A + cos A. 12. cot 8 A - cos 2 A = cos 2 A cot 2 A . 13. tan 2 A - sin 2 A = sin 4 A sec 2 A. 14. sec A +tanA = cosA/(l sin A). 15. (1 + sin A + cos A) 2 = 2 (1 + sin A) (1 + cos A). 16. cot 4 A + cot 2 A = esc 4 A esc 2 A. 17. (sin A + esc A) 2 + (cosA + sec A) 2 = tan 2 A + cot 2 A +7. 18. sin A (tan A 1) cos A (cot A 1) = sec A esc A. 19. tan 2 A - cot 2 A = sec 2 A esc 2 A (sin 2 A - cos 2 A). 20. 2 vers A + cos 2 A = 1 + vers 2 A. 21. cos 2 A (1 + tan 2 A) = 1. 22. (sec 2 A - 1) (esc 2 A - 1) = 1. 23. tan A + cot A = sec A esc A . 24. einA/cscA + cos A/sec A = 1. 25. sec 2 A -sec 5 A sin 2 A = 1. 26. (tan A - l)/(tan A + !) = (!- cot A)/(l + cot A). 27. (tan A + cot A) 2 = sec 2 A + esc 2 A . 28. (secA-cscA)/(secA+cscA) = (tanA-l)/(tanA+l). 29. (esc A - cot A) 2 = (1 - cos A)/(l + cos A). /*2 rt/2 30. Show that 2 + ^ = 1, if x = a cos A, j/ = 6 sin A. a o 78 THE TRIGONOMETRIC FUNCTIONS 49. Functions of angles at the quadrant limits : 1. Sin and cos 0. In Fig. 26 as OP moves toward OX, the angle, 0-0. At the same time the ordinate, y and x >h. We fire, therefore, justified in concluding 11 Lim sin = Lim f = r = 0. 9 >0 y 0'l n :. sin = 0. By similar reasoning, x h Lim cos = Lim-r = ,- = 1. e o* x*hh ft :. cos = 1. - i " i j- f^^ ' ' ' ' 'I A y\ Ojfcfc'~~ -jV \0 pt~~~ I \ ~" * / \ . Y' \ FIG. 26. In Fig. 26 OP = h, OP' = h', etc. 2. Sin 90 and cos 90 : By reasoning analogous to the above Lim sin = Lim \j = 77 = 1. sin 90 = 1. QUADRANTAL LIMITS 79 Again, x' Lim cos = Lim rj = -n = 0. g _> 90 z ' h h :. cos 90 = 0. 3. Sin 180 and cos 180: v" Lim sin0 = Lim rr/ = 777 = 0. 180 y"->otl n .-. sin 180 = 0. x" h" Lim cos 6 = Lim ^7, = -777- = 1. 0-+180" x"-^-^'^ ft /. cos 180 = -1. Note. It is evident that each of the limits above is the same whether the respective variables increase or decrease toward their limits. 1. Derive by a method similar to the above sin 270 = 1, cos 270 = 0; sin 360 = 0, cos 360 = 1. 2. From the values of the sines and cosines, by use of the fundamental formulas, obtain the values of the tangent, co- tangent, secant and cosecant of 0, 90, 180, 270, 360. 50. As was implied in 46, negative angles and their functions must be considered. It is easily seen that for every positive angle there exists a negative angle of equal magnitude. The coordinates of P' (Fig. 27) determine the functions of 6 in the same way that the coordinates of P determine the func- tions of 0. Now -y h Hence | sin 1 = | sin ( 0) | and (8) sin (-8) = -sin 8. Since x and h are the same for 9 as for 0, it follows that (9) cos (-8) = cos 8. 1. By use of the formulas of 47 derive tan (0), cot (0), sec (0) and esc (0) in terms of the same named functions of 0. 80 THE TRIGONOMETRIC FUNCTIONS 2. How can sin (35), cos (20) be determined from a table of sines and cosines where only positive angles are con- sidered ? Y FIG. 27. 51. If for any position of OP, Fig. 28, where P is the point P (x, y) a line OQ be drawn, where Q is the point Q (x\, yi), where x = yi> y = x i> then the angle XOQ will be the complement of the angle XOP. From the definitions of 47 and the above construction it follows that sin XOP = cos XOQ or sin a = cos (90 a), tan a = cot (90 - a), (10) and (11) and (12) sec a = esc (90 - a). These equations and the method of demonstration apply to all quadrants. These formulas will be named the complement relations. REDUCTION FORMULAS 81 1. From sin 30 = 0.5, find cos 60; From sec 45 = 1.414, find esc 45. 2. From sin 60 = 0.866, find cos 30; From tan 45 = 1, find cot 45. FIG. 28. 3. From cos 120 = -0.5, find sin (-30). 4. From tan 60 = V3, find cot 30. 5. The legs of a right triangle are x = 5, y 7. Calculate all the functions of the angle opposite the longest leg. 6. By use of formulas of 51 obtain all functions of the angle opposite the shortest leg from the values calculated in Ex. 5. 52. To reduce the functions of any angle to the same named functions of an angle less than 90. 1. Consider 90 < a < 180. Construct the triangles POM and P'OM', Fig. 29, so that h = h', a' = A, -x = x f , then y = y', A = (180 - a) = a'. Then sin a = y'/h' = y/h = sin a'. (13) /. sin a = sin (180 - a). 82 THE TRIGONOMETRIC FUNCTIONS Also, since x = x', (14) cos a = , = - = -cos (180 - a). FIG. 29. 2. Consider 180 < a < 270. Construct the triangles POM and P'OM' (Fig. 30) so that A = a r , y = -y', x = -x', then and a' = a - 180. Evidently, sin a = y/h = y'/h r = sin a/. (15) .'. sin a = -sin (a - 180). Also, cos a = x/h = x'/h r = cos a'. (16) /. cos a = - cos (a - 180). 3. Consider 270 < a < 360. Construct the triangles POM and P'OM (Fig. 31) so that A = a', y = y f , x = x', then h = h', and a' = 360 - a = A. Evidently, sin a = y/h = y'/h' = sin a'. (17) /. sin a = -sin (360 - a), and cos a = x/h = x/h' = cos a'. (18) .'. cos a = cos (360 -a ). REDUCTION FORMULAS 83 FIG. 30. FIG. 31. 84 THE TRIGONOMETRIC FUNCTIONS 4. Consider 90 < a < 180 and a 90. Construct the triangles POM and P'OM' so that y = x f , x = -y', then a' = a - 90. Then (20) Again FIG. 32. sin a = y/h = x'/h' = cos a'. sin a = cos (a 90). cos a = x/h = y'/h' sin a'. cos a = sin (a 90). FIG. 33. As an exercise let the student derive from these results the tangent of a, under all the above cases. ADDITION THEOREMS 85 5. Consider 360 < a. It is evident that if 360 be added to any angle a, the terminal line will coincide with that of a. It follows that the values of the defining ratios of the functions of a + 360 will be identical with those of a. It is evident then that (22) sin a = sin (360 + a), (23) cos a = cos (360 + a) , and similarly for all the functions. It is equally evident that 360 may be subtracted from a without affecting the values of the functions of a. 63. Addition theorems. The formulas (1) sin (A + B) = sin A cos B + cos A sin B, (2) cos (A -f B) = cos A cos B sin A sin B are known as the addition theorems for the sine and cosine respectively. Y FIG. 34. To prove (1), consider either position of Q (Fig. 34a and 346), where < A < 90 and < B < 90. Then A + B < 180. sin (A + fi) = NQ/OQ = MP/OQ + LQ/OQ. 86 THE TRIGONOMETRIC FUNCTIONS But MP = OP sin A and LQ = PQ cos A, since angle LQP A. Substituting these values in the above equation gives But sin (A + B] = -Qfi sin A + ~ cos A. OP , PQ = cos B and = am B. Substituting these values in the last equation gives (24) sin (A-\-B) = sin A cos B + cos A sin B. This is equation (1). Fia. 34. To prove (2) consider the same figure: i A m ON OM NM cos (A + B) = OM LP OQ OQ OQ OQ OQ But OM = OP cos A and LP = QP sin A. Substituting these values in the above equation gives: QP But cos (A + B) = ^ cos A ^~ sin A. OP PQ TTK = cos B and T^T\ = sin B. ADDITION THEOREMS 87 Substituting these values in the last equation gives (25) cos (A + B) = cos A cos B sin A sin B. This is equation (2). These proofs may be extended to angles of any magnitude by the use of 52. For a > 90, sin a can be expressed in terms of sin a' where a' < 90. Therefore, when A > 90 and B > 90 and A + B > 180 the sines and cosines of A and B can be expressed in terms of like-named functions of angles less than Fia. 35. 90, say A' and B', respectively. Corresponding to A + B > 180 there will be A' + B' < 180. Then sin (A + B) can be expressed in terms of sin (A' + B'). The generality of the equations (1), (2) for all angles is inferred. The proof may also be generalized. If 90 < A < 180, 90 < B < 180, 180 <A + B < 360. Then in the figure sin (A + B) = -sin (A' + B'} = y'/h' = -y/h, A + B - 180 = A' + B' = (A - 90) + (B - 90), where < A' < 90, < B' < 90, < A' + B' < 180. /. sin (A + B} = -sin (A' + B'), This case has been proved, since A' + B' < 180. 88 THE TRIGONOMETRIC FUNCTIONS It is known that: sin (A - 90) = -cosA = sin A', cos (A 90) = sin A = cos A', sin (B - 90) = -cos 5 = sin B r , cos (B - 90) = sin B = cos B'. From these tan(A+B)= -sin (A' .'. sin (A + B) = sin A cos B + cos A sin 5. This establishes the theorem for all values of A and B so long as A + B < 360. As an exercise let the student establish the equation (2) for the same values of A, B. In a similar manner the proof may be extended to the case of A -f B > 360. The addition theorems are general. The addition theorems for the other functions can be easily deduced from those of the sine and cosine by algebraic methods. Thus for tan (A + B), write , A . m sin (A + B) sin A cos B + cos A sin B tan (A + B) = . . , p ; = - = r 5 - cos (A -f- J5) cos A cos B sin A sin .B Dividing numerator and denominator of the last fraction by cos A cos B gives sin A cos B cos A smB / A i n\ cos A cos B cos A cos 5 tan (A + 5) = cos A cos B sin A sin B (26) /. tan (A + B) = cos A cos J? cos A cos 5 tan yi + tan B 1 tan A tan B for all values of A and B. 1. By use of the definitions of functions of negative angles derive from (24), (25), (26), (27) sin (A B) = sin A cos B cos A sin B. (28) cos (A B} = cos A cos B + sin ^4 sin B. /nr\\ / A tan A tan B (29) tan (A B) = r-r-r 77 & 1 + tan ^4 tan B THE SOLUTION OF TRIGONOMETRIC EQUATIONS 89 2. If sin A = 0.5 and sin B = 0.25, find sin (A + B). cos (A + B), tan (A + B), sin (A - ), cos (A - B), tan (A - B). Note. Cos A and cos B must first be found. Then sub- stitute in the addition theorems. 3. By use of the addition theorems derive: (30) sin (90 + A) = cos A. (31) cos (90 + A) = - sin A. (32) sin (90-^) = cos^. (33) cos (90-^) = sin,4. (34) sin (180 -A) = sin A. (35) cos (180 - A) = -cos A. (36) sin (A - 90) = -cos .4. 4. By use of the addition theorem for A=BorA+A = 2 A, derive (37) sin 2 A = 2 sin A cos A, (38) cos 2 A = cos 2 A sin 2 A = 2 cos 2 ,4 - 1 = 1-2 sin 2 .4. 5. By use of the addition theorems with A + 2 A = 3 A, derive expressions for sin 3 A and cos 3 A in terms of sin A and cos A. 6. Given sin A = 0.6, find sin 2 A, cos 2 A, sin 3 A, cos 3 A. 7. Given sin A = I , sin B = f , find sin (A + B) cos (A + B) tan (A + 5). 8. Given cos 2 A = 0.866, find cos A and sin A. M>fe. Write cos 2 4 = 2 cos 2 A - 1 = 0.866, and solve the equation for cos A. 9. Given tan 2 A = 1.5, find tan A and cos A. 10. Given tan A = 0.8, find tan (180 + A), tan (180 - A). 11. From sin 45 = 0.7071, find sin 22 30'. 54. In the solution of problems an unknown angle often occurs through one of its functions, say a sine or cosine. It is then necessary to solve the equation to obtain the angle. To do this the function of the unknown angle is regarded as the unknown quantity in the equation instead of the angle itself. 90 THE TRIGONOMETRIC FUNCTIONS When the function has been found the corresponding angle can be found from the table. To illustrate, suppose there is given 1 + sin A = f , from which to find A. Transposing and simplif ying give sin A = \. From the table, A is found to be 30. By use of formulas of 52 sin 30 = sin (180 - 30) = sin 150. Hence 150 is another value of A which satisfies the equation. Instead of finding A in degrees it may be desirable to use A merely as an angle belonging to its sine, |. For this purpose several notations are used. Thus A = angle whose sine is , A = arc sin \, A = sin" 1 ^, (read inverse sine ) all mean the same thing. Either of the equations A = arc sin x and x = sin A implies the existence of the other. Similar notations apply to all the functions of the angle. 1. What is the value of arc sin^? Of arc sin 1? Of arc Bin(-i)? 2. What is the value of arc tan 1? Of arc cos 0? Of arc sec 3? 3. What is the value of arc sin \ + arc cos | ? Of arc tan 1 + arc cot 1 ? 4. What is the value of sin (arc sin |) ? Of cos (arc sin I)? 5. What is the value of sin (arc sin \ -f- arc sin \ V) ? By the addition theorem this may be expressed as sin (A + B) = sin A cos B + cos A sin B, where A = arc sin \ and B = arc sin \ A/3. Then sin (arc sin (| \ + \ V3 \ V3)) = sin (arc sin 1) = 1. 6. If x = sin A and y = sin B, show that (A + 5) = arc sin (x Vl -y z + y Vl -a; 2 ). 7. Find x from arc sin 0.5 = x. GENERAL DIRECTIONS 91 8. Find x from arc tan 15 = x. 9. Solve for x in the equation arc sin x = arc cos (x ?). Note. Take the sine of both sides first. 10. Solve for x in sin (x - 25) = 0.6. Note. Expand the left side by the addition theorem, and use 48. 11. Solve for x in arc tan x = arc sec x 45. 12. Solve for x in arc tan x + arc cot x = 90. 13. Solve for x in tan (arc tan x) = 1. 55. One of the chief uses of the trigonometric functions is found in the solution of problems relating to triangles. All questions relating to right triangles can be solved by di- rect use of the definitions of go 47, for the first quadrant. For FIG 3g this purpose the definitions for acute angles can be modified as follows: Consider the A ABC in the figure, C being the right angle. Then y side opposite A smA = ~ = - , h hypotenuse x side adjacent A cos A = T = j h hypotenuse y side opposite A tan A = = - - j~ x side adjacent A Similar relations hold for the angle B. 56. General directions for solving problems relating to triangles: 1. Make a fairly accurate freehand diagram from the given conditions. 2. Mark all known measurements on the diagram. Note the position and relations of the unknowns. 3. Select a formula which will contain one of the unknowns and the knowns. 92 THE TRIGONOMETRIC FUNCTIONS 4. Substitute the values of the knowns in the formula and solve the resulting equation for the unknown. 5. When no formula fits the case directly, designate auxiliary unknowns by symbols. Formulate as many equations as unknowns and solve the system of equations simultaneously as in algebra. 1. Find the height of a tree, having given the angle of ele- vation, A = 35, and the distance AB = 125'. 125' FIG. 37. Note. The given measurements are the angle A and the side adjacent. The desired measurement is the side opposite A. We select, therefore, the tangent ratio. Write CB AB Substituting values From the table Solving for CB, tan A = tan 35 = 0.7002 = CJB 125' CB = 87.53. 2. From the ends of a line AB perpendiculars are dropped on MN meeting MN in C and D respectively. The angle GENERAL DIRECTIONS 93 between MN and AB is 47 30'. The line AB is 565 units long. Find CD. Fia. 38. Note. CD is called the projection of AB on MN. The angle AEB is called the angle of projection. See Geometry. 3. Find the projection of a line 50' long on a line at an angle of 60 with it. 4. What are the projec- tions on the axes of a line 120' long which is inclined 36 with the x-axis? The lengths of AB and CD in the diagram are wanted. 5. A plane surface may be projected on a plane in- clined to it in a manner exactly similar to the pro- jection of a line on another line. Find the area of the projection of a rectangle 30' x 60' on a plane inclined 75 to it. 6. A roof is 30' x 20' and is inclined 37 to the horizontal. How many sq. ft. of floor does it cover? 7. What is the area of the projection of a circle of 10' radius on a plane inclined to it at an angle of 27 ? 8. A circle has a radius of 10'. A chord of the circle is 15' long. How far is the chord from the center? 9. Find the perimeter of a regular pentagon inscribed hi a circle of radius 5". FIG. 39. 94 THE TRIGONOMETRIC FUNCTIONS 10. Find BD in the diagram if AC = 100', angle A = 25 and angle C = 30. D FIG. 40. Note. Introduce the auxiliary unknown x for CB and formulate two equations. Then eliminate x. 11. What is the eastward component of a force of 105 Ibs. acting in a direction 30 east of due north? The length OM rep- resents the eastward component. Find the northward component. 12. In the triangle ABC, A = 42, AB = 125, AC = 150. Find X the altitude on the side AC and the area of the triangle. From this problem determine a formula for the area of any triangle when two sides and their included angle are given. 13. By drawing an equilateral triangle and its altitude derive values of sin 30, cos 30, sin 60, cos 60. 14. By considering a square and its diagonal derive values of sin 45, cos 45. 15. A travels north 50 mi., then 37 east of north a distance 75 mi., then 10 west of south 125 mi. Find the length and direction of the line from the starting point to his final position. FIG. 41. THE SINE LAW OF TRIANGLES 95 Find the distance east or west he traveled and the distance north or south, from the starting point. 67. The sine law of triangles. From the figure it is seen that h = c sin A and h = a sin C. .'. c sin A = a sin C and sin C sin A FIG. 42. By drawing altitudes from the other vertices the equations b c and sin B sin C a b sin A sin B can be derived in exactly the same way as above. a b c (39) sin A sin B sin C These equations are known as the sine law. Its use is indi- cated when a side and the angle opposite are among the parts to be considered. 68. The cosine law of triangles. From the figure of the last section, c* = w + s 2 = h 2 + (b - t/) 2 = h 2 + 6 2 - 2by + y\ 96 THE TRIGONOMETRIC FUNCTIONS But h = a sin C, y = a cos C. Substituting in the above equation c 2 = 6 2 - 2abcosC + a sin 2 C + a 2 cos 2 C (40) or c 2 = a 2 + b 2 - 2 ab cos C. By using the other sides of the triangle as bases in turn, the following are derived in the same way: (40a) a 2 = b 2 + c 2 - 2 be cos A. (406) b 2 = a 2 + c 2 - 2 ac cos B. 59. Example of the use of the sine law. Given A = 15 19', C = 72 44', c = 250.4, of the triangle ABC, to find the remaining parts. Immediately B = 18G-(A ) = 91 57'. By the sine law, 250.4 _ sin 15 19' sin 72 44' 250.4 sin 15 19' sin 72 44' In logarithms this is log a = log 250.4 + tog sin 15 19' - log sin 72 44' = 2.3986 + 9.4218-10 11.8204-10 - 9.9800-10 1.8404 /. a = 69. 25. Again, to find b, using the sine law, sin 72 44' sin 91 57' Solving as above, 250.4 b 6 = 262.0. 1. Student check the result with natural functions, using the slide rule. EXAMPLE OF THE USE OF THE COSINE LAW 97 2. Construct the triangle to scale 50 to 1" from the given data and then measure the unknown parts. Check with the calculated values. 60. Example of the use of the cosine law. Given a = 1686, 6 = 960, C = 128 04', to find the other parts. By the cosine law c 2 = 1686 2 + 960 2 - 2 1686 960 - cos 128 4'.* Whence c = 2400. The values of A and B may now be found by the sine law. 1. Find AB, the distance across a river, from the data given in the diagram. C is a point on top of a hill, AB and CD are horizontal lines. BC = 500', angle DC A = 15, angle CBE = 20. River FIG. 45. 2. Two trains leave a station at the same time on straight tracks inclined to each other at an angle of 35. Train A travels 25 mi. per hr., train B travels 35 mi. per hr. How far apart are the trains at the end of 3 hrs. ? * Note that cos 128 4' is negative. 98 THE TRIGONOMETRIC FUNCTIONS 3. Find A B from the measurements given below. CD = 1000', ACD = 120, DCB = 30, ADC = 33, CDB = 105. 4. Use the cosine law to find the angles of the triangle ABC, if a = 75.8, 6 = 64.2, c = 81.9. 5. A tower stands at the top of a slope inclined 20 with the horizontal. From a point 800' down the slope from the tower the angle subtended by the tower is 5 25'. Find the height of the tower. 6. The tripod of a camera stands on a hillside. One leg is 3' long, the other two legs are each 5' and set on the ground at the same level. The three legs make equal angles with each other successively around. These angles are each 38. Find the distance between the feet of the legs. 61. Conversion formulas. The cosine law does not admit of use with logarithms. For this reason another law, derived from the sine law, is used when it is desired to employ logarithmic calculations. This law is known as the tangent law. Before the tangent law can be given some formulas must be developed. From the addition theorems we have: 1. sin (A + B) sin A cos B + cos .A sinB. 2. sin (A B) = sin A cos B cos A sin B. 3. cos (A -f J5) = cos A cos B sin A sin B. 4. cos (A B) = cos A cos B + sin A sin B. Add (2) to (1); subtract (2) from (1); add (4) to (3); subtract (4) from (3), and obtain the following equations in order. 5. sin (A + B) + sin (A - B) = 2 sin A cosB. 6. sin (A + 5) - sin (A - B) = 2 cos A sinB. 7. cos (A + B) + cos (A -B} = 2 cos A cos B. 8. cos (A + B) - cos (A - &) = -2 sin A sinB. P + Q In the last four equations substitute A = and * ' B = 7 ^. There results: THE TANGENT LAW 99 (41) sin P + sin Q = 2 sin P ^^cos P ~ ^ a & P -\- O P O (42) sinP-sinQ = 2cos ^ v sin g v . P 4- O P O (43) cosP + cos(> = 2cos jp-cos a a P 4-O P O (44) cos P - cos Q = -2 sin ;. sin 62. The tangent law. By the sine law a _ sin A b sinB Taking this proportion by composition and division gives a 4- b _ sin A -j- sin B a b sin A sin B Applying (41), (42), to the right side gives, by use of 48, fAK\ a + 6 (45) r = Two similar equations using the other parts of the triangle are obtained in a similar manner. This equation is called the tangent law for triangles. This law applies directly to the case that was solved by the cosine law in 60. That problem will now be solved by the tangent law. Write 1686 + 960 tan \ (180 - 128 04') 1686 - 960 " tan \(A - B) since \ (A + 5) = \ (180 C) in any triangle. Solving the above equation for tan \ (A B), 726 tan 25 58' Applying logarithms to this equation, log tan i (A - 5) = log 726 + log tan 25 58' - log 2646. Substituting the values log tan HA - B) = 9.1258 - 10. Whence | (A - B} = 7 37'. 100 THE TRIGONOMETRIC FUNCTIONS Now %(A + B)+%(A-B) = A = 25 58' + 7 37' = 33 35', and %(A+B) -%(A-B) = B = 25 58' - 7 37' = 18 21'. Having the angles A and B, the remaining side can be found by the sine law. 63. Ambiguous case. When the given parts of a triangle are two sides and an angle op- posite one of these sides any one of the following results may occur: Let a, c, A be the given parts and let h be the altitude from B. 1. a < h, where h = c sin A. Under these circumstances the triangle cannot be constructed. 2. a = h, where h = csin A. There is, under these circum- stances, one triangle, a right triangle with the right angle at D. 3. a > h and a < c, where h = csin A. Under these con- ditions there will be two triangles, ABC and ABC'. Note. C' + C = 180. Hence sin C" = sin C. In solving this case the value of C will be obtained from its sine. Since there are two angles each less than 180 having the same sine, a choice must be made between C' and C. This choice must be based on other data in the problem. When no conditions are given for making a choice both solutions must be given. Note. Whenever in solving a problem a result is obtained that implies that the sine or cosine of an angle is greater than 1, the problem is either impossible or an error has been made in the calculations. (a) Example. Given a = 250, A = 42 12', c = 600, to find the remaining parts of the triangle ABC. By the sine law, sin C sin 42 12' 600 250 Solving gives log sin C = 10.2075 -10. DOUBLE AND HALF-ANGLE FORMULAS 101 This is equivalent to sin C > 1, which is impossible. This means 250 < h and the triangle cannot be constructed. (b) Example. Given c = 254.3, a = 396.8, A = 94 29', to find the remaining parts of the triangle ABC. By the sine law 396.8 = 254.3 sin 94 29' sinC' Solving, log sin C = 9.8054 -10. /. C = 39 43' or 140 17'. Since an obtuse angle occurred in the given data, only the smaller value of C can be used. (c) Example. Given a = 250, c = 300, A = 42 12', to find the remaining parts of the triangle ABC. By the sine law, sin C = sin 42 12' 300 = 250 Solving, log sin C = 9.9064 -10. .-. C = 53 43' or 126 17'. K b *i In this case either value of C will satisfy the problem and there are two possible triangles. The other parts of each triangle can be found. 64. Double and half-angle formulas. In the addition theorems put B = A. Then (46) sin (A + A) = sin 2 A = 2 sin A cos A. 102 THE TRIGONOMETRIC FUNCTIONS 47. cos (A + A) = cos 2 A = cos 2 A - sin 2 ,4 = 2 cos 2 A - 1 From 47, 2 sin 2 A = 1 - cos 2 4 ,.. . . /I cos 2 A or (1) sin A = y - - From 47 again, 2 cos 2 A = 1 + cos 2 A or (2) cos A = In (1) and (2) put A = -% and obtain: (48) sinlp: (49) cos 1 /1 + cosP 2 r V 2 As an exercise derive tan \ P, cot \ P, sec \ P, esc | P. 65. Angles of a triangle in terms of its sides. By the cosine law: 52 g2 _ a 2 (1) Add 1 to both sides, 6 2 (2) l+cosA=- Subtract both sides of (1) from 1, a 2 (b 2 2 be + c 2 ) _ (a-j-b c) (a b+c) Substituting these values in (48), (49), 64, gives: (K.f]\ eini A Wl sin 75/1 (51) cos l, RADIUS OF CIRCLE 103 where a + 6 + c = 2s. Note the arrangement of letters in the last two equations. A is any angle of the triangle: 1 . Derive tan ~ A from (48) , (49) . Tan ~ A = V r-^ r - z & \ a) 2. Find the angles of the triangle ABC it a = 45, 6 = 55, c = 66. 66. Radius of circle inscribed in a triangle ABC. Call a + 6 + c = 2s. By the geometry of the figure: A2 + 53 + Cl = s or A2 = s - 3 - Cl = s - 53 - C3 = s a. Now or (52) r = (s a) where a is any side and A the opposite angle of the triangle. 67. Radius of circle circumscribed about a triangle ABC. In Fig. 49, angle A' = angle A, angle A BC = 90, 104 Therefore and (53) THE TRIGONOMETRIC FUNCTIONS . , . , BC BC a sin A = sin A' = - TT7l = ^-5 = ' - TTl -5 -5 A'C 2R 2R a ^ -r 2 sin A where A is any angle of the triangle and a the opposite side. 68. Circular measure of angles. We are accustomed to measure angles in degrees. It is often convenient to measure angles with another unit. This unit is called the radian and is denned by the equation, (54) BicAB = r -9, where r is the radius of the arc AB and is the angle in radians. Since an arc equal to the circumference has the same measure as a 360 angle at the center the relation between the degree and the radian is easily obtained. FIG. 50. CIRCULAR MEASURE OF ANGLES 105 By (54), 9 = in radians, that is, the arc divided by its radius gives the measure of the subtended angle in radians. Therefore (55) and arc (circumference) _ 2irr _ r r 2 TT radians = 360. 360 1 radian = 27T = 57 17' 45"- The following table of equivalents is easily derived and will be used in work to follow. It would be well to memorize this table. Radians. Degrees. 2x 360 IT 180 7T/2 90 7T/3 60 r/4 45 1T/6 30 1. The radius of a circle is 12. The angle at the center is 7T 45 = radians. Find the arc intercepted on the circumference. By (54), arc= 12- = ST. 2. Find the arc subtended by a chord 4' from the center of a circle of radius 10'. The cosine of half the angle subtended by the chord is T % = 0.4. From this the angle is found from the tables to be 23 35' or 0.41+ radians. Now by (52), arc = 10 X .41+ = 4.1+, ft. 3. In a circle of radius 12" a line 10" from the center is drawn. What is the length of the arc cut off? (First find the angle subtended by the chord, then apply Eq. (52.)) 106 THE TRIGONOMETRIC FUNCTIONS 4. The radius of a circle is 15, a chord is drawn cutting off a segment of altitude 3. Find the area of the segment. 5. A carriage wheel is 4' in diameter, the carriage is traveling 10 mi. per hr. Find the number of revolutions per min. and the number of radians per sec. turned through by the wheel. 6. The arc through which a pendulum swings is 4". The length of the pendulum is 39.4". Find the angle of swing in radians and in degrees. 7. An arc is 10" and the angle measured at the center is 1| radians. Find the radius of the circle. 8. A horizontal tank 6' in diameter and 30' long is filled to a depth of 2'. Find the number of gallons in the tank. See Chap. II (51). 68a. Mil, a unit of angular measure: Just as a central angle standing on an arc that is -5^ of the circumference of a circle is one degree, so also a central angle standing on an arc that is s 5 Vrr of the circumference is one mil. Since by this definition there are 6400 mils at the center of a circle, it follows that the length of the arc that subtends one mil is circumference _ 2irr _ 6.283 r _ mftQQ1 ~ 6400 ~ 6400 ~ "6400" : 1. In military work it is common to speak of a mil as ap- proximately equal to an arc of one foot at a distance of 1000 feet or one yard at a distance of 1000 yards. Give reason for this statement. 2. In a table of natural sines of angles, expressed in mils, is the following: Angle in mils 64 1600 2400 3180 sine .0628 1 .6071 .0194 Check this table by converting mils to degrees and then use table of natural sines. 3. From problem 2 find the sine of 800 mils. 69. Explanatory definitions. (a) In surveying the direc- tion of a line is usually given by giving the angle which it makes with a north and south line. If the line runs north and east its PROBLEMS 107 direction is designated by N. 6 E., where 6 is the angle between a line running due north from the starting point and the line of sight from same point. If the line runs in a northwesterly direction its direction is designated by N. 6 W. Similar notations apply to directions southeast- erly and southwesterly. (6) The angle of ele- vation of a point is the angle between a horizon- tal line and a line from the observer to the ob- served point, above the level of the observer. (c) The angle of de- pression of a point below the observer is the angle between a horizontal line and a line from the observer to the point observed. In the figure, 9 is the angle of elevation of B from A, and tf is the angle of depression of C from A. PROBLEMS. 1. The distance between two points measured on a slope of 5 42' with the horizontal is 210.3'. Find the horizontal distance between the points. 2. The distance between the points A and B measured on the horizontal is 388.0'. The bearing from A to B is N. 30 E. How far is B north and east of A? 3. A surveyor sets his instrument over a stake at A and reads the bearing to another stake at B, S. 22 10' E. The distance from A to B is 1142.1' measured on an upward slope of 12 21'. Find how far B is north or south of A. How far is B east or west of A? How far is B above or below A? (Save results.) 4. Compute similarly the position of C with reference to B when the bearing from B to C is S. 37 30' W. and the distance 843.7' measured on a downward slope of 10 25'. (Save results.) 5. Using the results of (3) and (4) compute the position of C with respect to A. FIG. 51. 108 THE TRIGONOMETRIC FUNCTIONS 6. It is desired to drive a straight passage from A to B in a mine. A is known to be 3500' north and 2200' west of a third point C. B is known to be 2500' south and 600' west of C. Find the length and bearing of the straight passage from A to B. 7. A horizontal line 1033 ft, is measured in the same vertical plane with the top of a mountain. From one end of the line the angle of elevation of the top of the mountain is 13 22', from the other end the angle of ele- vation is 5 10'. What is the height of the mountain above the horizontal line? 8. On the right bank of a river, two stakes A and B are set 250.0' apart on a horizontal line. On the left bank a stake C is set so that the angle A in the triangle ABC is 90. The angle B is measured to be 38 41'. What is the distance from A to C ? 9. From a hill top the angles of two points on opposite sides of the hill are 20 33' and 15 10', respectively. The distance from the hill top to the first point is 1200 yds. The distance to the second point 2000 yds Find the distance between the points. 10. A surveyor took measurements as follows: AB = 500', angle DAB = 100, angle DAC = 67 45', angle ABC = 125, angle DBG = 70. Find the length of DC. 11. A flag pole stands on a tower 50' high. From A, on a level with the base of the tower, the angle of elevation of the top of the tower is 42 35'. From A the pole subtends an angle of 10 15'. Find the length of the flag pole. 12. Find the difference between the areas of the triangle, ABC, and its inscribed and circumscribed circles, if the sides of the triangle are a = 56.3, b = 76.5, c = 68.8. 13. What are the angles of the triangle whose sides are, a = 665, 6 = 776, c = 887? 14. Two stones are one mile apart on a straight road, From a tower standing on the roadside the angles of depression of the stones are 15 and 3 50', respectively. Find the height of the tower and its distance from each stone. Note. There are two cases: (a) when the tower is between the stones; (6) when the stones are on the same side of the tower. 70. Graphic representation of the trigonometric functions. (a) Construct the graph of the equation y = sin x. To do this, the values of the angle x must be laid off in linear units to some scale. From 68, if r = 1, the arc = the angle at the center expressed in radians. This furnishes the unit of GRAPHIC REPRESENTATION 109 measure, viz., an arc equal in length to the radius. Lay off values of x along the re-axis to the scale of 1 radian = 1". Tabulate values of x, and y = sin x below. z y = sin x. X y = sin i. = radians*.. 0.0 210 = 7 ir/6 radians.. -0.5 30 = 7T/6 0.5 225 = 5x/4 -0.707 45 = x/4 0.707 240 = 47T/3 -0.866 60 = x/3 0.866 270 = 3T/2 -1.0 90 = 7T/2 1.0 300 = 5x/3 -0.866 120 = 27T/3 0.866 315 = 7 7T/4 -0.707 135 = 3 x/4 0.707 330 = llT/6 -0.5 150 = 5 ir/6 0.5 360 = 2 T 0.0 180 = 7T 0.0 ,2! 2L 6 |3 <-r=l-H 137T FIG. 52. A property of trigonometric functions known as periodicity is illustrated by this curve. This was suggested in 52. It is noticed that as a point traces the above curve the values of y are the same for positions of the point going in the same direc- tion on the curve, such as the pairs of points A, B; A', B'; A", B". It follows that corresponding to any value of sin x, there are infinitely many values of x differing successively by 360 or 2 TT radians. This fact may be expressed by sin x = sin (x n 360) = sin (x 2 rnr), * Note that T = 3.1416, then ir/6 radians = 0.5236 radians and similarly with other values. 110 THE TRIGONOMETRIC FUNCTIONS where n = 0, 1, 2, 3, ... A similar relation holds for each of the functions cos x, tan x, cot x, sec x, esc x. It may be further noted that there are two equal ordinates of the curve y = sin x in each of the intervals to 180; 180 to 360; 360 to 540, etc. From this fact it occurs that the value of a single function alone, say of sin x or tan x, is not suffi- cient to determine which of the two values of x between and 180 or between 180 and 360 is to be taken. Either another function must be known or else the quadrant in which the angle lies must be known. This fact was illustrated in the solution of the ambiguous case of triangles, 63. 1. Construct the graph of y = cos x in a manner similar to that employed in constructing the graph of y = sin x, above, j 2. Construct the grapk of y = sin x + cos x. Note. Tabulate values of sin x and cos x and add the corre- sponding values. Use these sums as ordinates of the curve. 3. Construct the graph of y = sin 2 x. Note. Erect the ordinates equal to sin 2 x, over the values of x. 4. In a manner similar to that indicated in Ex. (3), construct the graph of y = sin 3 x. How do the curves in this exercise and the preceding differ from the graph of y = sin x ? 5. Construct the graph of y = sin ^ x. 6. Construct the graph of y = tan x. What peculiarity has this curve at x = ir/2, x = 3 ir/2, etc.? 7. Construct the graph of y = - sin x, given that the value dp ,. r ., /sinx\ of limit 1 = 1. *-o \ x I 8. Construct the graph of y = sec x. What peculiarity does this curve have at ir/2, 3 7r/2, etc. ? In what way does it differ from the sine curve and the tangent curve at x = 0, x = TT, etc. ? 71. The equation y = arc sin x may also be studied by means of its graph. This function is infinitely many valued as was shown in 52. GRAPHIC REPRESENTATION 111 Values of y = arc sin x for different values of x. 0.5 0.866 -1.0 -0.5 -0.866 -1.0 360, 720, . . . 27T, 47T, . . . 30, 30 360, . . . 7T 7r o 6' g27T, ... 150, 150 360, ... ^ *-/B STT STT 137T ^/ T' 6 2, ... -g- -/B 60, 60 360, ... ,/ 117T if "" O f~* 6 3' 3 ^ 27r > /_ 22L 3 120, 120 360, ... / 3T 2 IT " 2 \ 2 :r- 2?r . V 47T 3 3 \T 3 90, 90 360, . . . \-- ~6~ 7T 2 2' I*!..,.:, > \ 5ir 210, 210 360, ... 6 . 2ir \ 7T i 7T A T 1 O "" \ '2 ~~j 330, 330360, ... ^ jr.' UTT [ O _. 6 ' 6 '< r-i T / oAft Z4U , 47T 4x AV~- 7T M - i 3 ' 3 :b2T, . . . ^T 1 6 300, 300 360, ... Fie i. 53. 270, 270 360, . . . 37T 37T T 2 Zir, . . . 7T / TT The figure shows the graph for values of y from ^ to + -5-- o o 112 THE TRIGONOMETRIC FUNCTIONS The curve shows that any line parallel to the y-axis and less than a unit distant from it cuts the curve in many points, that is, there are many ordinates for each abscissa. Such a function as y = arc sin x is called a multivalued function of x. The smallest positive ordinate for any value of the abscissa is called the principal value of the ordinate or angle. 1. Construct the graph of y = arc cosx. 2. Construct the graph of y = arc tan x. 3. Construct the graph of y = arc sin 2 x. 4. Construct the graph of y = arc sin J x. 72. Equations involving trigonometric functions as unknowns are of frequent occurrence. They may be solved in much the same way as algebraic equations. 1. Find the principal value of x that will satisfy the equation 2 sin 2 x 3 cos x = 0. This can be written, 2 (1 cos 2 x) 3 cos x = 0. Solving for cos x, cos x = 0.5 or 2. /. x = arc cos 0.5 or arc cos 2. The first gives x = 60 and 300, of which 60 is the principal value. The second is impossible since | cos x \ = 1. This value must, therefore, be rejected. 2. Solve and determine the principal value of x in 2 sin 2 z cos 2 x = 0. Note. First remove the double angle by substituting its equivalent from formulas of 53. 3. Solve cos 2 e sin 6 = %. 4. " 4 cos = 3 sec x. 5. " 3sin0-2cos 2 = 0. 6. " tan 6 + cot = 2. 7. " 4sec0-7tan 2 = 3. SPECIAL INTERPOLATION 113 Note. It may happen in solving an equation containing two functions, that eliminating one function will lead to only part of the solution. The other part of the solution may be obtained by eliminating the other function. 72a. Tables requiring special interpolation. Gradients are commonly called grades or slopes and are expressed: (a) By the angle which the line of direction makes with a horizontal line. Example. A gradient of 1, 2, 3, etc. (6) By the change of elevation corresponding to a given horizontal distance. Example. An elevation of 4' in 100', that is, a 4 per cent slope. An elevation 1' in 60' gives a gradient of ^V (read: 1 on 60). Below is a table which gives differences of elevation for gradients of to 5, and horizontal distances. Gradient in degrees. Difference of elevation for horizontal distance of: 1 2 3 4 5 6 7 8 9 0.5 1 2 3 4 5 0.0087 0.0175 0.0349 0.0524 0.0699 0.0875 0.0174 0.0350 0.0698 0.1048 0.1398 0.1750 0.0261 0.0525 0.1047 0.1572 0.2097 0.2625 0.0348 0.0700 0.1396 0.2096 0.2796 0.3500 0.0435 0.0875 0.1745 0.2620 0.3495 0.4375 0.0522 0.1050 0.2094 0.3144 0.4194 0.5250 0.0609 0.1225 0.2443 0.3668 0.4893 0.6125 0.0696 0.1400 0.2792 0.4192 0.5592 0.7000 0.0783 0.1575 0.3141 0.4716 0.6291 0.7875 Find the difference of elevation for a gradient of 3, and a horizontal distance of 567 ft. From table, on line with gradient of 3, For distance 7 the elevation is 0.3668 For distance 60 = 6 X 10 elevation is 3 . 1440 For distance 500 = 5 X 100 elevation is. . 26.2000 Total 29.7108 For distance 567 ft., therefore, elevation is 29.71 ft. 114 THE TRIGONOMETRIC FUNCTIONS EXERCISES. 1. Check the above method of interpolation by using your table of logarithms and the formula: elevation = horizontal distance X tangent of the angle. 2. Make out a table corresponding to the table above replacing hori- zontal distances by distances measured along the slope, using the formula: elevation = distance along slope X sine of angle. 3. Compare this table with one in text. What conclusion do you draw when the angle is small ? In solving problems connected with traverse sailing or with land surveying, it is often necessary to find the latitude and departure corresponding to the several courses and distances. Since latitude = distance X cos of bearing, departure = distance X sin of bearing, it is convenient to use a table in which these projections have been computed. Following is a section of such a table: TRAVERSE TABLE. Dist. 1. Dist. 2. Dist. 3. Dist. 4. Dist. 5. O 8f> On rw Lat. Dep. Lat. Dep. Lat. Dep. Lat. Dep. Lat. Dep. 30 15' 0.8638 0.5038 1.7277 1.0075 2.5915 1.5113 3.4553 2.0151 4.3192 2.5189 59 45' 30 8616 5075 7233 0151 5849 6226 4465 0302 3081 5377 30 45 8594 5113 7188 0226 5782 5339 4376 0452 2970 5565 15 31 8572 5150 7142 0301 5716 5451 4287 0602 2858 5752 59 15 8549 5188 7098 0375 5647 5563 4196 0751 2746 5939 45 30 8526 5225 7053 0450 5579 5675 4106 0900 2632 6125 30 45 8504 5262 7007 0524 5511 6786 4014 1049 2518 6311 15 32 8480 5299 6961 0598 5441 5S98 3922 1197 2402 6406 58 15 8457 5336 6915 0672 5372 6008 3829 1345 2286 6681 45 30 8434 5373 6868 0746 5302 6119 3736 1492 2170 6865 30 45 0.8410 0.5410 1 6821 1.0810 2.5231 1.6229 3.3642 2.1639 4.2052 2.7049 15 33 8387 5446 6773 0893 5160 6339 3547 1786 1934 7232 57 Dep. Lat. Dep. Lat. Dep. Lat. Dep. Lat. Dep. Lat. C/OUTS6. CoUTSG. Dist. 1. Dist. 2. Dist. 3. Dist. 4. Dist. 5. From this table determine the latitude and departure of a Bourse with bearing 31 10' and length (distance) 413. SPECIAL INTERPOLATION 115 Determine the latitude and departure of a course with bear- ing 31 10' and length 413. A surveyor runs a line N. 30 45' E., distance 243, then N. 32 15' W., distance 214. How far north and how far east or west is he from the starting point? By use of the traverse table, find the area of a right triangle which has a hypotenuse 534 units in length and an acute angle 59 10'. A table of logarithmic and natural haversines is indispen- sable in connection with problems in nautical astronomy. Be- low is a portion of such a table. LOGARITHMIC AND NATURAL HAVERSINES hav 6 = % vers 6 = $ (1 cos 6) = sin 2 6/2. a. 3" 50 m 57 30'. 3 h 51 m 57 45'. 3 h 52 m 58 0'. 3 53 m 58 15'. 3 h 54 m 58 30'. a. Log hav. Nat. hav. Log hav. Nat. hav. Log hav. Nat. hav. Log. hav. Nat. hav. Log. hav. Nat. hav. 9.36427 .23135 9.36772 .23319 9.37114 .23504 9.37455 .23689 9.37794 .23875 60 1 .36433 .23138 .36777 .23322 .37120 .23507 .37461 .23692 .37800 .23878 59 2 .36439 .23141 .36783 .23325 .37126 .23510 .37467 .23695 .37806 .23881 58 3 .36444 .23144 .36789 .23329 .37131 .23513 .37472 .23699 .37811 .23884 57 +1' 9.36450 .23147 9.36794 .23332 9.37137 .23516 9.37478 .23702 9.37817 .23887 56 5 .36456 .23150 .36800 .23335 .37143 .23519 .37484 .23705 .37823 .23891 55 6 .36462 .23153 .36806 .23388 .37148 .23523 .37489 .23708 .37828 .23894 54 7 .38467 .23156 .36812 .23341 .37154 .23526 .37495 .23711 .37834 .23897 53 +2' 9.36473 .23160 9.36817 .23344 9.37160 .23529 9.37501 .23714 9.37840 .23900 52 20" 9 20 h 8 m 20" 7" 20" 6 m 20" 5 m Solve the following problems by use of above table (do not determine 6} . Log hav 6 = 9.37144-10, find nat. hav 6. Nat. hav 6 = 0.23893, find log hav 6. t (time) = 3 h 51 m 3 s , find log hav t and nat. hav t. t = 20 h 7 m 56 s , find log hav t and nat. hav t. Find log hav and nat. hav of 58 0' 25". 116 THE TRIGONOMETRIC FUNCTIONS 1. The angle at the center of a circle is 4 radians. The arc is 7'. What is the radius of the circle? 2. What is the value, hi degrees of -=- radians? Of -=- radians? i O 3. What is the value in radians of 40? 55? 1200? 4. An arc is 5'. The radius is 8'. What is the angle at the center in radians? In degrees? 5. One angle exceeds another by -=- radians. The sum of the angles o is 175. Find the angles. 6. Prove sin a cos a = sin 3 a cos a + cos 3 a sin a. 7. Prove cot a esc a = I/ (sec a cos o). 8. Prove (1 + tan 2 a) /(I + cot 2 a) = sin 2 a/cos 2 a. 9. Prove sin a ski o/(cot a 1) = sin a [1 sin a/ (cos a sin a)]. 10. Given tan 30 = $ Vjj, find tan 15, by use of a formula. 11. Using the addition theorem find sin (a + b + c). 12. Prove cos 2 a = (1 - tan 2 o)/(l + tan 2 o). 13. Prove sin f j a 1 = (cos a sin a)/ V2. 14. Prove tan a + tan b = sin (o + 6) /cos a cos &. 15. Prove cos a = (1 - tan 2 i o)/(l + tan 2 1 a). 16. Prove cos (45 + a) /cos (45 - a) = sec 2 a - tan 2 a. Solve and determine all values less than 360. 17. tan a = sin a. 18. (3-4 cos 2 a) cos 2 a = 0. 19. sin a + cos a cot a = 2. 20. tan (45 + o) = 3 tan (45 - o). 21. 5 ski a = tan a. 22. 2 cos a + sec a = 3. 23. 2 sin a + 5 cos a = 2. 24. tan 2 a = 1. 25. If tan 2 o = m, find tan o. 26. If cos o = J, find cos J a. 1 ^ 27. Prove arc tan ^ +arc tan c = T> given tan -. = 1. 28. Solve for y in 2 arc sin | + arc sin y = w/2. 29. In the figure the following data are given to find AB. CD = 943.4; CE = 673.3; a = 72 9.3'; ft = 60 17.9'; y = 32 14.6'; 5 = 67 33.9'; e = 19 14.7'. See Fig. 54. Ans. AB = 1054. 30. A coal mine entry runs N . 1 8 W . to A whose coordinates are (450, 100) . A post on the boundary line has coordinates (575, 475) . The bearing of the MISCELLANEOUS EXERCISES 117 boundary line is N. 26 50' E. Another entry has been run N. 80 E. to a point B across the boundary whose coordinates are (112, 174). It is desired to continue the first entry to a point 10' from the boundary, then parallel to the boundary to a point where the second entry would intersect it if continued, FIG. 54. then to follow a line that would coincide with the "continuation of the second entry. Make the calculations and determine thejcoordinates of the points of turning. 31. Find the coordinates of P from the following: Start at A whose co- ordinates are (-75, 350); run S. 26 15' W., 355'; then run S. 54 20' E., 175'; then run N. 10 15' E., 300'; then run N. 75 45' W., 500' to P. 32. Since the mil is the angle at the center of a circle subtended by sdrs of the circumference, therefore, Measure in mils of angle A _ 6400 _ 160 Measure in degrees of angle A 360 9 Change the following angles in degrees to mils: 45, 90, 135, 180, 225, 270, 315, 360. By means of the definition of a mil and that of a radian determine the conversion factor for changing angles from mils to radians. Change the following angles in radians to mils: ir/2, ir/4, if. Change the following angles in mils to radians: 800, 1600, 2400, 3200. 33. By the definition of a mil, if the radius of a circle is 1000 yds., an arc of 1 yd. subtends an angle at the center of approximately one mil. In 118 THE TRIGONOMETRIC FUNCTIONS the figure, therefore, from similar triangles, assuming / and arc s equal, s 1000 rki* o = R 1000 or s = -5- F, F R or, since the angle t in mils has the same measure as the arc s, 1000 t = H F. Note. The assumptions made in the above discussion give results sufficiently accurate, in general, for work in gunnery where t is usually a very small angle. (o) If the range R = 1000 yds., and the distance between the two guns F = 20 yds., what is t in mils? G, -1000- FIG. 55. What is t if (6) R = 1000yds., and F = 40yds.? (c) R = 1000 yds., and F = n yds. ? (d) R = 2000 yds., and F = 20 yds. ? (e) R = 2500 yds., and F = 30yds.? 34. The angle at the target T (or aiming point P) subtended by the distance between two successive pieces G and G\ on the battery front is called the parallax of the target (or aiming point). If the aiming point is in the line of the battery front the parallax of the point is a minimum, zero. If the target (aiming point) is on the normal to the battery front, GGi at the mid-point the parallax is a maximum. This is called the normal parallax. In practice, if the target is on any normal to the battery front, In what follows the normal through FIG. 56. its parallax is considered as normal. G is chosen. MISCELLANEOUS EXERCISES 119 The normal parallax t of T is (see exercise 33, above) GG l F /e/iooo where R is the distance from T to battery front. (a) What is the normal parallax of the target for a range 2500 yds., and distance 20 yds. between G and GI? (6) What is the normal parallax if R = 500 yds. and F - 40 yds.? 35. If the target is not on the normal to the battery front, a correction for "obliquity" is required. This is called "correction of parallax due to obliquity." The angle at G between the line normal to the battery front and the line to the target T' is the angle of obliquity. In the figure, is the angle of obliquity of the target T'. The true parallax (t) is the normal T parallax after it is corrected for obliquity. A """^"Vr' True parallax is obtained as follows: DG = GiG cos DGGi, approximately, But t' (mils) = DG ^ . (See Exercise 33.) Therefore t' (mils) = = t (mils) cos 0, 1000 ** 1 G GiG =- = normal parallax of T. FIG. 57. Therefore, true parallax = normal parallax X cosine of the angle of obliquity. (o) If the normal parallax t = 8 mils, and the obliquity angle is 600 mils, find the true parallax. (6) If R = 2500 yds., the obliquity angle 600 mils, and G,G = 20 yds., find the true parallax. (c) Verify the formula, true parallax t' = r ' ^L , where angle is as K/ \\j\j\j indicated in the figure. CHAPTER IX POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 73. The position of a point can be determined by means of a distance and an angle. Let OX be a fixed reference line, called the axis. Call the pole. Then if the angle 6, and the distance OP = r, are known the position of P is known. The point P is designated as P (r, 6) or as (r, 0). The number pair (r, 0) are the polar coordinates of P. OP = r is the radius vector of P, and is the vectorial angle of P. FIG. 58. If P lies on the terminal line of the vectorial angle, r is reckoned as positive. If P lies on the terminal line of the vectorial angle produced on the opposite side of the pole, r is regarded as nega- tive. The vectorial angle 6 is positive or negative according as it is reckoned counter clockwise or clockwise, respectively. In the figure, if P' is regarded as P' (r'd"), r' and 0" are positive. If P' is regarded as (r', 6'), r' is negative. If P' is regarded as 120 POSITION 121 (r'0'"), r' is positive and 6'" is negative, and similarly in other cases. 1. Locate the following points (4, jj, (6,30), [5, o)> 2. Construct graph of r = 8 6 (6 in radians). Note. Arrange a table of values as in drawing graphs on rectangular coordinates. Then lay off the points as above and draw a smooth curve through the points. Assume = 0, TT 6' FIG. 59. 7T 7T 3' 2' Calculate rv = 0, o g, 8-3, A portion of the graph is shown in the figure. 3. Construct the graph of r = 8 cos 0. 4. " " 5. " * 6. " * n (i <i 8. " " 9. " " 10. r = 2/(l cos0). r = sin 2 6. r sin 6 = 4 (solve for r). r = 4/(l -3cos0). r = 7 (a circle). r = cos 3 0. r = cos - 11. In 2 x + 3 y = 5, put x = rcosd,y = rsind and draw the graph of the resulting equation. Draw the graph from the original equation. 12. In y 2 = 8 x, put y = r sin 0, x = r cos + 2. Note the form of the resulting equation. Construct the graphs of both, equations. 122 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 74. In the solution of certain quadratic equations and equations of higher degree, there occur roots of the form a + bi, where a, b are real and i 2 = I or i = V 1, 36gr. From the above definition of i it is easily found by calculating that, i v i, ^ = (V^I) 2 = -1, # = #.{ = (_i) V^T = -V^i = -i, The value of i 5 is the value of i, the value of i 6 will be the value of i 2 , etc. 75. A geometric basis for interpreting complex numbers is attributed to Argand. Since any number or any line segment has its sign or direction reversed when multiplied by 1 = i 2 , the line is rotated 180 counter clockwise by multiplying by -1 = i 2 . Thus in the figure AB - (-1) = AB V = -AB = AB'. It then looked reasonable to suppose that AB is rotated 90 counter clockwise when multiplied by V 1 = i or AB is just half reversed. Thus AB i = AB" in the figure. This idea proves to be very useful. The number a + bi may now be looked upon as a point in the plane determined by two steps taken perpendicular to each other. Thus, if a = Om and b = mn in Fig. 61, a + bi will be Om + mP, since bi is perpendicular to 6. This idea is then equivalent to regarding (a, 6) in a + bi as the rectan- gular coordinates of a point P. Polar coordinates may also be associated with P. Thus, if angle XOP = 6, and OP = r, P may be regarded as P (r, 6). The line OP = r is called the modulus or absolute value of FIG. 60. ARITHMETIC OPERATIONS WITH COMPLEX NUMBERS 123 the number a + bi. The angle 6 is called its amplitude, the figure it is easy to see that for any position of P, From (1) (2) (3) (4) r = Va? + b 2 . sin0 = - r COS0 = - r tan 6 = a FIG. 61. These relations are funda- mental. 1. Locate on rectangular coordinate paper the points, 2 + 3 i (o = 2, b = 3); 2- 3i; 1 4i; 4 3i; 5 + 2i. 2. Find the modulus and amplitude of each number in (1). 76. Arithmetic operations with complex numbers. (a) Two complex numbers are equal when and only when they represent the same point referred to the same axes. Hence the two complex numbers a + bi and a' + b'i are equal when and only when a a' and b = b'. (b) The sum of two complex numbers a + bi and a' + b'i is the complex number a + a' + (6 + b') i. Thus the sum of 3 + 2i and 1 + 4i is 4 + Qi. 1. Add 3 + 5 i to 1 + 2 i; I + 6 i to 3 - 2 i; - -$ i to ? + $ i, locate each point and each sum on a diagram in rec- tangular coordinates. 2. Add ( -2 + 3 i), ( - 1 - 2 i), (3 + 6 i). Locate all points and the sum. 3. Add (1 + 3 i), (-3 - i), (6 + 7 i). Locate all points and the sum. (c) Multiplicatibn of two complex numbers is defined by (a + bi) (a' + b'i') = aa' - bb' + (ab' + a'b) i. Note. Remember in carrying out the work, i 2 = 1. 124 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS (d) Division of complex numbers is defined by a + bi (a + bi) (a' - b'i) = aa' + W (a'b - ob') i a' + b'i (a 1 + b'i) (a' - b'i) a'* + b' 2 H a' 2 + b' 2 The operations (6), (c), (d) are in general possible and lead to complex numbers, impossible if a' 2 + b' 2 = 0, indeterminate if a* + b' 2 = a 2 + b 2 = 0, see 42, 43, 77. 1. (3-4i)(7 + 4i) =? (3-4i)(-7-4i) = ? Locate all points and the products. 2. (3 + 4i)-5-(7 + 4i) = ? (3-4i)-r-(-7-4i) = ? Locate all points and the quotients. 3. (2 + 3i)-5-(l-r) = ? (l-i) (l+i)-5-2 + 2i = ? Locate all points and the quotients. 77. By the formulas of 73, a + bi may be put in the form a + bi = Vtf+W-=J= + . U \Va 2 + b 2 Va 2 + b 2 = r (cos0 + isin0). The last form is called the polar form of the complex number (a + bi), where r = Va 2 + b 2 , sin 6 = . and cos = f CVJ.J.V4. \J\JYJ I/ / 'a 2 + b 2 Va 2 + b 2 Multiplication and division of complex numbers take very interesting and useful forms in polar coordinates. Thus (a + bi) (a' + b'i) = r (cos d + i sin 6) r' (cos tf + i sin 00 = rr' [(cos 6 cos 0' - sin sin 0') + i (sin cos 0' + cos sin 0')] = rr' [cos (0 + 0') + i sin (0 + 0')]. This shows that the modulus of the product is the product of the moduli and the amplitude of the product is the sum of the amplitudes of the numbers multiplied. Again, a + bi' r (cos + i sin 0) (cos 6' i sin 0') a' + b'i ~ r' (cos 8' + i sin 0') (cos 0' - i sin 0') = -, [(cos (0 - 0') + i sin (0 *- 0')]. T This shows that the modulus of the quotient is the modulus of VECTORS 125 the dividend divided by the modulus of the divisor and the amplitude of the quotient is the amplitude of the dividend minus the amplitude of the divisor. As an exercise reduce each complex number in the last section to polar form and perform the indicated operations by use of the above formulas. 78. There are a number of physical quantities, of fundamental importance, such as force, velocity, electric field, etc., that are completely specified by magnitude and direction of the line of action. They are called vector quantities in distinction from non-directed quantities, such as energy, speed, etc. The latter are called scalar quantities for the reason that they are com- pletely specified by magnitude alone. A directed straight line segment is a vector. It is evident that a vector may represent a vector quantity. For the length of the vector, to some Y scale, may represent the magnitude of the vector quantity and the vector may be parallel to the line of action of the vector quantity. If A B is a vector, its direction is understood to be from A toward B. A is the initial point of the vector AB, and B is the terminal point. A vector may be dis- placed parallel to itself without affecting its mag- nitude or direction. When a vec-tor may be so moved it is called a free vector. When a vector is attached to a fixed point, it is a fixed or localized vector. 79. The notation of complex numbers and of polar coordi- nates lends itself readily to representing vectors and to cal- FIG. 62. 126 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS culating with them. Thus the vector OP in Fig. 62 is de- scribed either as (x + iy) or as (r, 6) at pleasure. Since a vector symbolizes a vector quantity, any operations of arith- metic performed with vectors will have a similar meaning with vector quantities. The direction of OP is that from to P. When OX is the reference line (axis) the angle 6 determines the direction of OP. 80. Addition and subtraction of vectors. The most con- venient notion from which to derive vector addition and subtraction is displacement or step. Thus to add a vector BC = b* to another vector AB = a, lay off a and then from the FIG. 63. terminal point of a lay off b. The vector from the initial point of a to the terminal point of b, that is, AC = c, is the vector sum of a and b. This may be expressed by the vector equation a + b = c. That is, a displacement AB a followed by the displacement BC = b is equivalent to the displacement AC = c. If the vector B'C' is to be subtracted from the vector AB', lay off the vector B'C" equal in length to B'C' but in the oppo- site direction. The vector AC" is the difference. If AB' = a'j B'C' = b' and AC" = c' we may write the vector equation a' - b' = c'. * When a single letter represents a vector heavy type is used. ADDITION AND SUBTRACTION OF VECTORS 127 It is seen that to subtract a vector is the same as to add its opposite or negative. To add several vectors OP, OP', OP", ... lay off the vectors successively, with in- itial point of each on the terminal point of the preced- ing, forming the sides of a polygon. The closing side of the polygon drawn from the starting point to the terminal point of the last vector is the vector sum of the given vectors, Fig. 65. The vector equation is OP" + OP + OP' = O'P'. FIG. 64. FIG. 65. It must be remembered we are not adding the lengths of the lines alone but the displacements (including directions of the lines). The sum of two sides of a triangle is vectorially equal to the third side, but not numerically. 1. Find the vector sum of two adjacent sides of a parallelo- gram ABCD. Find the vector difference of the same sides. 2. Find the vector sum of the three sides of a triangle ABC, 128 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS taken in order. Find the vector difference AB BC. Find the vector sum AB + BC. 3. Find the sum of the vectors whose initial points are at the pole and whose terminal points are (3, 60) ; (10, 20) ; (24, 120), respectively. 4. Find the sum of the vectors represented by (3 6 i) ; (4 + 2t); (5-3t); (l-4t). 5. Find the magnitude and direction of the velocity resulting from two simultaneous velocities, one due north 50'/sec., the other N. 45 E.,30'/sec. Note. By 75, 76 these velocities may be expressed in rec- tangular form and added. The sum can then be reduced to polar coordinates. The result may be found by use of the law of 58. Solve by both methods and check the results. 6. A boat is rowed across the current of a river at 5 mi./hr. The current is 1| mi./hr. Find the actual velocity of the boat in magnitude and direction. 7. If a horse is running N. 35 E. at the rate of 12 mi./hr., how fast is he going north ? How fast east ? 8. Make diagrams to scale for Exs. 6, 7 and verify your calculations. 9. Can you explain by the vector idea why a division of society into opposing factions retards or prevents social progress? 10. Show by vector addition the truth of the parallelogram of forces. 81.* Product of Two Vectors. Vectors exhibit two types of product: (a) Scalar product of two vectors is defined as the product of their magnitudes and the cosine of their included angle. Thus if F = r (cos e + i sin 0) (77, 79) and S = * The remainder of this chapter may be omitted if desired. It is recommended, however, if the time permits and the student's knowledge of mechanical notions justifies, that the entire chapter be covered carefully. PRODUCT OF TWO VECTORS 129 the scalar product of F and S will be written as FS = rr' cos (0' - 0) = FS cos (0' - 0). Since cos ( a) = cos a, the order of taking the factors is in- different. (6) Vector product of two vectors is defined as the product of their magnitudes and the sine of their included angle. The vector product of F and S will be written as vFS and is numeri- cally equal to FS sin (0' - 0)* = rr' sin (0' - 0), where 7^, S are the magnitudes of F, S, respectively. As the name indicates the vector product is a vector. The direction is that of the travel of a right-hand screw, perpendicular to the plane of the two vectors, when turned in such a way as to rotate the first factor toward the second. Then 0' is regarded posi- tive. If the direction of rotation is re- versed the angle 0' becomes negative and its sine becomes negative and con- sequently the vector product changes sign and its vector is reversed. The order of factors in the vector product is, therefore, not indifferent, but must be carefully noted. 1. Find the scalar product of (3, 40) and (6, 30). Find the vector product, factors taken in order given. Note: 30 -40 = -10. 2. Find the scalar and vector products of (25, 35) and (60, 124) in order given. 3. If work is defined as the product of force multiplied by the displacement component in the direction of the force, show that work is the scalar product of force and displacement. * It is noted that 8 sin (&' 0) is the perpendicular distance from the origin to a line through the terminal of S parallel to F* This distance, in case .F* is a force, is called the moment arm of F about the origin as an axis. See Ex. 5 below. 130 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 4. What is the work done by a force of 2000 Ibs. acting N. 30 E. in moving a car 50' on a track due north? (No friction.) 5. The moment of a force about a point or axis is defined to be the product of the force by the perpendicular distance of the line of the force from the point or axis. Show that the moment of a force about any axis is the vector product of the vector distance from the line of the force to the axis and the vector force. Note the order of factors. 6. Find the moment of a force of 500 Ibs. acting N. 40 E. about a point such that the vector from the point to the initial point of the force vector is 12 units in a direction N. 80 E. 7. The coordinates of the initial point of a force vector are (3, 6), the terminal point (4, 10). Find the moment of the force about the origin. The magnitude of the force is the length of the line joining the above two points. The moment arm is the distance of the line from the origin. 8. Find the scalar product of (16, 30) by (24, 60). 9. Find the vector product of the vectors in Ex. 8. Note. We shall, from now on, describe a vector having its initial point at the origin by giving the coordinates of its ter- minal point only. 82. A vector may be re- garded as the vector sum of its components on the axes. Thus the vector OP is the vector sum of Om and mP. We shall adopt the notation of writing a sub- script to indicate the com- ponent. Thus the vector OP will be denoted by P and its component on OX and its component on OF by P y . The vector equation o FIG. 68. holds for all vectors. P = VECTOR PRODUCT 131 The scalar and vector products of two vectors can now be expressed in rectangular coordinates. If F = F^ + F V} S = So, + S v we may write: (1) FS = FA + F V S V . Now if F is a force and S a displacement then F, is a force and Sg, a displacement in the same direction and FJS,,, is the work done by FO, in the displacement So,,. Similarly F V S V is the work of F y in the displacement S y . The right side of (1) then is the measure of work done by the components of F. But work is the product of force and the component of displacement in the direction of the force, ''that is, the product of force and dis- placement and the cosine of their included angle. There- fore, the scalar product on the left of (1) represents the same thing that the right side repre- sents and the two members of (1) are but different ways of expressing the scalar product of F and S. (2) vSF = S X F V - S V F^ a vector. The term S m F v is the moment of the force F y acting at A and tending to swing A counterclockwise about 0. The term S v Fa. is the moment of F x acting at A tending to swing A about clockwise and is negative. Strictly speaking the right side of (2) is a scalar quantity, but owing to the known conventions as to the direction of rotation it contains the means of determin- ing the direction of the tendency to rotation and we are justified in calling it a vector. The left side of (2) is known to be the moment of F about when applied at A, 81. Hence the two sides of (2) represent the same thing. We infer the generality of (1), (2), for all vectors. These products play an important role in physics and mechanics. FIG. 69. 132 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 1. Find the axial components of (35, 26); (125, -65); (-275, -120). 2. Find the sum of the components along the z-axis and the sum of the components along the y-axis in (1). Consider these sums as components of a new vector and calculate its modulus and amplitude (75). 3. Find the modulus and amplitude of the vector whose components are Pa, = 496, P v = 275. 4. Find the moment of P, = 50, P v = 75 applied at the point (10, 8). 5. If F = 30 + i 50 and S = 8 + 7 i, find the scalar and vector products of F and S. 6. Reduce (25, 120) and (80, 40) to rectangular coordinates and find the scalar product and the vector product. 7. In (6) find the scalar and vector products without reduc- ing to rectangular form. 8. Find the total moment about the origin of the vectors (6 + 10 i), (-4 + 3 i), applied at the point (-3, 7). 9. A lever has weights i An as shown in the diagram. Find the distance from A to the point of application of a 100 Ib. force that will just balance the other forces. Note. The sum of the v - n moments of the downward IG. / U. forces about A must equal the moment of 100 Ibs. about A where x is the unknown moment arm. That is, lOOz = 5 50 + 11 100 + 19 25, to find x. 83. The notation of vectors will now be applied to some problems in the equilibrium of particles and of rigid bodies acted upon by external forces. r- "T EQUILIBRIUM OF PARTICLES 133 A particle is a geometric point regarded as having inertia or mass. A rigid body is one of finite fixed magnitude and unvarying form. (a) Equilibrium of a particle. In order that a particle shall not have its state of rest or of uniform motion in a straight line changed, all forces acting on the particle must be balanced. That is, there must be no component of resultant force along any line or axes of reference. Since the forces may be repre- sented by vectors, this means the vector sum of all forces acting on the particle must be zero. This in turn means the polygon formed by the vectors in succession must be a closed polygon. This polygon is called the vector polygon of forces. If the vector sum of the forces acting on a body is not zero there will be a resultant force equal to the closing side of the polygon which is the vector sum of the vectors of the forces. The calculation of the resultant force or vector sum of a set of forces is an important problem. (6) A rigid body is in equilibrium when it is at rest; moving uniformly in a straight line, without rotation; rotating uni- formly about a fixed axis or moving uniformly in a straight line and rotating about an axis whose direction is fixed. For equilibrium of a rigid body (a) the conditions of equilib- rium of a particle must be satisfied; (6) the sum of all moments acting on the body must be zero, when taken about any axis. The solution of a problem relating to a rigid body is generally in two parts, viz. : First, consideration of the forces as acting on a particle of the same mass as the body; second, determination of the moments and the resultant moment acting on the body. Some examples will illustrate how problems in equilibrium of particles and rigid bodies may be solved in ordinary cases. I. Let a particle m be acted on by the forces P, Q t R, all in the same plane as shown in Fig. 71. To find the resultant force. P = 75, Q = 100, R = 125. The vector equation for the resultant is: Resultant = S = R + + P. 134 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS To calculate the magnitude and direction of the resultant, resolve along two perpendicular axes, mX and mY. Along mX, S x = S cos 6 = R cos 120 + Q cos 10 + P cos ( -20) or = 125 (-0.5) + 100 (0.985) + 75 (0.939) = -62.5 + 98.5+70.4. /. S x = 106.4. T FIG. 71. Along mY, S v = S sin 6 = 72 sin 120 + Q sin 10 + P sin (-20) = 125 (0.866) + 100 (0.173) + 75 (-0.342) = 108 + 17.3 - 25.6. .'. S, = 99.7. = V& 2 +S V 2 = V106.4 2 + 99.7 2 = 141.3 S 99.7 Now and = 0.938. .-. 6 = 43 10' (see figure, ms = S). In a similar manner an unknown force in any system acting on a particle can be determined if the equations of equilibrium can be written. If the vectors are drawn carefully to scale the value of S can be found from the vector polygon. The method of construc- tion can be seen from Fig. 65, 80. Start at m and lay off the vectors in order. The closing side S is the resultant. This method of solution is called the graphic method. EQUILIBRIUM OF PARTICLES 135 A third method, called the geometric method, is as follows: Draw the diagram and solve the problem by methods of Chap- ter VIII. See Figs. 72, 73. First find angle a from the known directions of R and P. Then AC can be found. Having AC, determine angle BAG and angle CAO. Now OC and angle /3 can be found. Let the student carry out the work. It is strongly recommended that every problem be solved by two of the three methods. The graphic method may very well be used as one in each case. II. Consider the case of a trap door as shown in the figure. It is desired to find the pull (tension) in the rope and. the hinge reaction in the position given in Fig. 73. Fa FIG. 73. FIG. 74. It is seen at first that the weight of the door regarded as applied at its center of gravity, G, tends to produce clockwise rotation about A, the hinge. Further the rope applied at B tends to produce counterclockwise rotation about A. The measure of tendency to produce rotation by a force is the moment of the force. For equilibrium of the trap door the sum of the moments about A must equal zero. That is, the 136 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS two moments must be equal but in opposite directions, that is, of opposite sign. These two moments must be calculated. The moment due to the pull, T, in the rope, is equal in mag- nitude to M = T - AB sin DEC (a magnitude of the vector product) = T-AB sin CBA = T 10 sin 70 = 9.39 T (counter clockwise). The moment of the weight of the door is M z = w - AGsinCAG (magnitude of vector product) - 100 5 . sin 60 = 433 (clockwise). /. 9.39 T = 433 and T = 46.1 Ibs., the tension in the rope. To find the hinge reaction at A, we find its horizontal and vertical components. By taking horizontal components of all forces, calling the horizontal component H, T co&EBC + H - cos = 0, 46.1 cos 140 + H = 0, whence H = 35.4. By taking vertical components, T sin EEC + V sin 90 + W sin 270 = 0, 46.1 sin 140 + V - 100 =? 0, whence V = 70.4. Now if P is the hinge reaction P = V(70.4) 2 + (35.4) 2 = 79.1, approximately, 704 tan 6 = ^2 = 2, nearly, = 63 25', approximately. Thus the problem is completely solved. 1. The forces shown in the dia- gram act on a particle. Calculate the magnitude and direction of the Ex. 1. resultant. Note. Take horizontal and vertical components as in 1. EQUILIBRIUM OF PARTICLES 137 2. Find the force P required to hold the lever AB in position as shown in the figure. -:*- 51 \ Ex. 2. 500 3. Find P so that the load at B will be sustained, CAB being a bent lever with fulcrum at A. Ex. 3. 4. Find a force P making an angle 50 with OX so that the system will be held in equilibrium. Determine also the unknown angle, 6. Ex. 4. 5. Three men carry a heavy uniform bar weighing 450 Ibs., one man at one end, the other two with a short stick some distance from the other end. If the length of the bar is I, find how far from the end the two must lift so all lift the same amount. Bar 450 Ex. 5. 6. Find the supporting ^ forces at the ends of the I horizontal bar loaded as shown in the figure. 100 -9-- 300 Ex. 6. 138 POLAR COORDINATES, COMPLEX NUMBERS. VECTORS 7. Find the stress in the rope of the crane shown in the figure. Find the forces Vi, Vz, con- sidering ABC as a single rigid body acted upon by Vi, V 2 and 2000 Ibs. Note. Start by taking resolutions at (7, then mo- ments at A. 8. Three boys pull on three ropes tied to a ring as shown in the figure. Determine the pulls in ropes AB, AC. (Resolu- tions.) 9. Determine the forces in the frame as shown in the fig- ure. 10. What pull on the rope is necessary to hold the boom in position with the load as indicated in the figure. 11. Find the resultant of two forces of 500 Ibs. each, one acting due north and the other N. 60 E. Ex. 7. Ex. 9. Rope Ex. 10. CHAPTER X EQUATIONS 84. Let/ (x) be defined by the expression: / (x) = a& n + aiz"-' + azx n ~ 2 -f + a n -ix + a n , where n is an integer and the a's are known constants. A function of this kind is called an integral function of the variable x, of degree n. Theorem I. If r is a root of the equation: /(*)-0, then / (x) is exactly divisible by x r. For divide/ (x) by (x r) by the ordinary method, continuing the process until a remainder not containing x is obtained, the result can be represented as (1) f(x-) = (x-r)q(x)+R, where q (x} denotes the quotient and R the remainder. If r is a root of/ (x) = 0, the left side vanishes for x = r. The first term on the right also vanishes for x = r. Then (1) becomes (2) = + R whence R = 0. Therefore, the division is exact and (3) f(x) = (x-r)q(x). 85. Assumption. Every integral equation * has at least one root. Theorem n. Every integral equation of degree n has n roots. Write the equation as (4) /(*)=0. * That is an integral function equated to zero. 139 140 EQUATIONS This equation has a root, say r\. By Theorem I: (5) /(*) = ( -r,)/! (a?) =0, where f\ (x) denotes the quotient. Now as above (6) /i(*)=0. If /i (x) is not a constant it is an integral function of x of degree n 1, and Eq. 6 has a root, say r 2 . Hence (7) /i(*) = (*-*)/(*). Continue this process until a quotient f n (x) not containing x is obtained. No more divisions can be carried out and no more roots exist. The result is that / (x) has been broken up into factors as shown in the equation, (8) f(x) =a Q (x- n) (x - r 2 ) . . . (x - r) = 0. The values r\, r 2 , . . . , r n are the only values of x which satisfy this equation and consequently the only roots of equation (4). Incidentally equation (8) shows how to form an equation that shall have given roots. 86. Theorem HI. If the equation / (x) = has more than n roots it is an identity and has infinitely many roots and every coefficient is zero. Consider, (9) / (x) = aoz n + aix n ~ l + '+ ctn-is + a n = and suppose it has more than n roots. If every a is not zero the terms whose coefficients do not vanish will form an equation of degree not higher than n which can have not more than n roots. But this contradicts the hypothesis. Hence the theorem is true. This theorem has important uses in mathematics, some of which will appear in the sequel. Determine which are identities and which are not. 1.] (x a) 2 = z 2 2 ax + a 2 , expand left side, transpose and collect terms. 2. x 2 4 = 0. For how many values of a; is this equation satisfied? ZERO AND INFINITE ROOTS OF INTEGRAL EQUATIONS 141 .2 Q 3. r- ~ x 3. Clear of fractions. x + 3 4. r 5 - 3 x 2 - x = 0. 5. 1 -X ' 1 -3 6. Is 2 a root of x 2 + 4 z + 4 = 0? Is -2 a root? 7. Is 1 a root of z 3 - 3 z 2 + 3 x + 1 = 0? Is 2 a rootl 87. Theorem IV. If f (x) be divided by (x r), the remainder will be / (r). For write / (x) = (x r) q (x) + R. Now put x = r f (r) = (r r) q (r) + R- Sincer-r = 0, R=f(r). This theorem furnishes a convenient method of calculating f (r). To shorten the work it will be desirable to learn an abbreviated method of performing the division by x r. This will be given later. 88. Zero and infinite roots of integral equations. Con- sider (1) f(x) = aoz n + ajx"- 1 + + a*-\x + a n - 0. If a n = a n -i = - = a n -jt+i = 0, k roots of (1) are zero. Under this hypothesis (1) may be written, (2) a& n + Oix"" 1 + + a n -ix + a n = x k (a& n - k + atf 1 -*- 1 + + fln-t) = 0. It is seen now that k roots are zero since x k = (x 0) fc is a factor of the expression constituting the left member. Now substitute x = - in (1) and obtain, after multiplying by y n , (3) a n y n + a n -iy n ~ l + + a# + a<> = 0. If in (3) OQ = ai = 02 = = a k -i = 0, k roots are zero, as shown above. But by virtue of the relation, x = -, x = oo y when y 0. Therefore, there is an infinite root of (1) for each 142 EQUATIONS zero root of (3). Hence when the first k coefficients of (1) approach zero, k of its roots become infinitely large. Zero roots often occur in practical work, and infinite roots have important meanings in certain types of problems. 1. What is the value of x in xy = k, when y = 0? If this equation is regarded as the equation of a curve what is the interpretation for y = 0? 2. What is the value of y = x 2 2 x + 1, when x = 1? Interpret this result as in Ex. 1. 89. Synthetic division is a shortening of the process of long division when the divisor is a binomial of the first degree in the variable. Consider the example: x 3 -5x z \x z + x -2 x 2 7 x -2s + 10 If the cancelled terms are omitted the work appears as -5x 2 \x* + x - 2 x 2 -7x -5x + 10 By pushing up the remainders into a line under the dividend the work appears as - 5 x 2 - 5 x + 10 - 2 x The x's may be omitted, using coefficients only. Thus 1-4-7 + 10 |-5 -5-5 + 10 |l + l-2 1-2 THEOREM V 143 It is noticed that the last form gives all the information given in the first form of division. If the coefficient of the first term of the dividend is brought down, then the partial remainders hi order are the coefficients of the quotient and the final re- mainder is zero. In applying this method it is desirable to use addition instead of subtraction during the process. This is done by changing the sign of the second term of the divisor, that is change 5 to 5. Thus, 1-4-7 + 10 [5 + 5 + 5-10 1+1-2 90. Theorem V. If / (x) is exactly divisible by x r, r is a root of the equation / (x) 0. This theorem is proved by means of the equation used to prove Theorem I. For if R = 0, f(x) = (x -r}q(x'). Substituting r for x causes the right side to vanish, and hence the left side also. Therefore, r is a root of / (x) = 0, by defini- tion of root. It is now easy to use Theorem V and synthetic division to determine whether or not any given number is a root of a given integral equation. If a number should not be a root it is useful to know that the final remainder R is the value of / (r). In applying synthetic division, if any term in is missing its place must be filled with a zero. Solve by synthetic division: 1. Is 2 a root of x 3 - x* - 3 x - 3 = 0? 2. Is 1 a root of z 3 - z 2 + 3 Z - 3 = 0? 3. Is x - 3 a factor of 3 z 4 - 11 x 3 + 5 x 2 + 3 x = 0. 4. Factor 6 x 2 + 19 x + 10. 5. Factor x 3 + 5 x z + 2 x + 10. 6. Factor x*-2x* - 8 z - 16 (supply term in z 2 by 0). 144 EQUATIONS 7. Find any integral root of x 3 25 x z + 8 x 16 = 0. 8. Find any integral root of 2 z 3 - 3 z 2 3 z + 2 = 0. 9. Find any integral root, of z 4 + 2 z 3 5 z 2 4 z + 6. 91. Solution of numerical equations in one unknown of any degree. No general simple rule can be given for finding the roots of equations of degree higher than the second. It is possible to solve equations of the third and fourth degrees by formulas, but the application is in general not easy. For practical purposes some method of approximation is most useful. Such a method can be easily constructed from the preceding theorems. For values of x not roots of / (x) = 0, it is obvious that the result of substituting such values in / (x) would give positive or negative values of the function instead of zero. For values of x near r, where r is a root of / (z) =0 and h a small positive number, one of the following relations must, in general, hold: (a) /(r-*)</(r)=0</(r + ft), or (6) /(r-/0>/(r)=0>/(r + A), or (c) /(r-ft)</(r) =0>/(r + A), or (d) /(r-*)>/(r)~0</(r + *) l or (e) /(r-ft)=/(r)=0=/(r + ft). Conditions (a) and (6) are the ones which will now be con- sidered, being the most commonly occurring. These can be used to discover the position of the real roots of an equation of any form. Consider first an integral equation, /(z) = z 3 -4z 2 -2z + 8 = 0. Note. What follows illustrates a method. In practice one woif d first determine whether integral factors of 8 are roots. Try, by synthetic division,* in succession the values 3, 2, 1, 0, 1, 2, 3, etc.: * Synthetic division is here only a convenient method of substituting values of x in the equation. SOLUTION OF NUMERICAL EQUATIONS 145 1-4- 24- 8 1-3 -3 + 21-57 -7 + 19-49 -2 + 12-20 x = -3, R =/(-3) = -49 x= -2,R=f(-2) = -28 x = -1, #=/(-!) = +5 By condition (a) a root lies between 2 and 1. = +8 x= l,R=f(l) = +3 x = 2,R=f(2) = -4 By condition (6), a root lies between 1 and 2. 1-4- 2+ 8 2- 4-12 -2- 6- 4 = -7 'r = 4 7? = f 61^ = *( *T , J i I \~tj vl i 4isaroot. 0-2 Bringing down the first coefficient the quotient is x 2 2. Now having found one root, 4, exactly the equation can be written x 3 - 4 x 2 - 2 x + 8 = (x - 4) (x 2 - 2) = and the remaining roots are easily found from the equation x 2 - 2 = 0. But to illustrate the method, when an exact root is not found the fact that 4 is a root will be ignored and we shall proceed to find the root that lies between 2 and 1. 146 EQUATIONS Take a value of x midway between 2 and 1, that is 1.5, 1-4 -2 +8 | -1.5 s= -1.5. fl=/(-1.5)= -1.375 - 1.5 + 8.25 - 9.375 By condition (a) a root lies - 5.5 + 6.25 - 1.375 between -1.5 and -1. Now take the value midway between 1.5 and 1. Since this value involves three figures, it will be better to take -a near value in two figures as 1.2 or 1.3. Again, since the value of R for x = 1.5 is smaller, numerically, than the value of R for x = 1, we will risk choosing 1.3. 1-4 -2 +8 1-1.3 689-636 *= -1.3, fl = / (-1.3) = 1.64. l.O T^ O.OW O.OD T- j. 1 -I r 1 1 n _ i . -_ , ., - . Root between 1.5 and 1.3. o.o + 4.o9 + 1.64 Now take 1.4, midway between 1.5 and 1.3. 1-4 -2 +8 LlL* z= -1.4, fl=/(-1.4) = +0.216. - 1.4 + 7.56 - 7.784 Root between _ IA and _ L5> - 5.4 + 5.56 + 0.216 Since /( 1.4) is much smaller (numerically) than /( 1.5) we will try values quite near 1.4. Take 1.41. 1 -4 -2 +8 - 1.41 + 7.628 - 7.935 - 5.41 + 4.628 + 0.065 Take -1.42, 1_4 _2 +8 -1.42 + 7.696-8.088 Root^between -1.41 and - 5.42 + 5.696 - 0.088 The root is, therefore, 1.41 correct to three figures. The nearly equal values of R for 1.41 and 1.42 would suggest that the next figure is near 5 and a continuation of the work will show that the next figure of the root is 4 and the root is 1.414 correct to four figures. This value is sufficiently exact for all ordinary purposes. The above process is rather laborious. Practice will enable one to obtain three or four figures of a root quite readily. None r , ~"' /( > SOLUTION OF NUMERICAL EQUATIONS 147 of the current methods of finding irrational roots are much shorter, if any. Some are longer and require more theoretical knowledge. 1. Find the roots of z 4 + x 5 - 7 x* - 5 x + 10 = 0. 2. Find the roots of . 28z 4 + 239z 3 + 1020z 2 + 813z-140 = 0. 3. Find the root of Ex. 13, following 34, by the above method. Consider another type of equation, x sin x 1 =0. Synthetic division cannot be employed here as a method of substituting values of x, for the reason that this equation, as will be shown later, has an infinite number of terms when we expand it into an integral equation. Such an equation is called transcendental. In this case the angle, x, in the first term must be given in radians. The values of sin x are to be taken from a table of natural sines. Try x = 2 radians = 114 35', nearly. 2 - 0.9095 - 1 = +0.0905, x = 2 rad., R=f(2) = 0.0905. If x = 1.9 radians = 108 51' 19 09463-1- -00463 -- =/ (1-9) = -0.0463. l.y U.y^rOO 1 U.vrrUO, I -r> j. I. ir in i- J Root between 1.9 and 2 radians. If x = 1.95 radians 1.95 - 0.9291 - 1 = 0.0209, x = 1.95 rad., R = f (1.95) = 0.0209. If x = 1.93 radians 1.93-0.9362-1= -0.0062, x = 1.93 rad., #=/(!. 93) = -0.0062. This value is correct within an error of less than angle of 10'. Further trials will give more accurate results. Solve by repeated substitution : 1. x = tanx. 2. x = 3 sin a; 1. 3. 3 x + logio x = 5. 4. sinz + cosx = 1.4. 148 EQUATIONS Note. In some cases it may be useful to construct the graph of the equation before attempting the solution. The intercept of the graph on the axis of abscissas will furnish a guide in selecting trial values for the root. 5. The distance from the earth of a body projected vertically upward is given by the formula s = v t- 16.1 1 2 , where v is the velocity of projection upward in feet per second and t is the time from starting, in seconds and s is the distance in feet. If a body is projected upward with a velocity of 80'/sec., in how many seconds from starting will it be 30' from the earth? (Two values of t.} 6. A stone is dropped into a well. It is heard to strike the bottom after 5 sec. Having given s = 16.1 t 2 and the velocity of sound = 1150'/sec., determine the depth of the well. 7. The volume of a cubical box is diminished 1200 cu. in. by putting in it 1/2 in. lining on all faces. Find the original dimensions of the box. 8. The compound amount of $3000 at x per cent, for 5 yrs., is $3500 (x Y 1 + 155] 9. Find the length of a chord of a circle that cuts off \ its area if the radius is 1. 10. Find the x-mtercepts of the curve whose equation is y = 2x s -z 2 -6z + 3. Note. Write the function equal to zero and solve the equation for its roots. 11. How deep will a sphere of wood 1' in diameter sink in water if the density of the wood is 0.7 that of water, given the volume of a spherical sector equals the area of its spherical surface times ^ its radius and the volume of a cone equals | its base times its altitude. 12. Find the real fifth root of 15. PARTICULAR CASE OF QUADRATIC EQUATION 149 Note. x 5 15 = 0. Solve this equation in the regular way. 13. In the solution of a certain problem in mechanics the result depended on solving the equation cos 3 + 0.8 cos 2 6 - 0.02 cos - 0.393 = 0. Determine one root to three figures, (cos as unknown.) 92. Particular case of quadratic equation. The frequent occurrence of equations of the second degree (quadratic equa- tions) makes it desirable to give them some special treatment. All the theorems regarding integral equations, given in this chapter so far, hold for quadratic equations. By Theorems II, III a quadratic equation has two and only two roots. It can, therefore, be written in the form (1) k(x- n) (x - r,) = 0, where k is constant and r t , r 2 are the roots. Equation (1) can be written (2) kx* -k(ri + r 2 ) x + km = or x z (TI + r 2 ) x + rir 2 = 0. The equation (2) is of the form (3) az 2 + bx + c = or z 2 + -z + - = 0. a a Comparing the left-hand members of (2) and (3) it is easily seen that if they represent the same equation b c TI + TZ = and rir 2 = a a The formula given in 10 may be obtained as follows: Given ax z + bx + c = 0. Dividing by a, . 6 c x* + -x = a a 150 EQUATIONS b z Adding -. 5 to both members. 4 a 2 ,.2_L& ,_&!_ *>* c_ *"** 1 ~4a 2 ~4a 2 a~ 4a 2 Taking the square root of the first and last members, _6_ " or x = 2a -6=b\ / 6 2 -4ac 2a The two roots of the quadratic will be equal to each other if 6 2 4 ac * = 0. In this case each root is 6/2 a. What will be the nature of the roots if 6 2 4 ac < 0? What will be the nature of the roots if 6 4ac>0? By calculating the expression 6 2 4 ac it is possible to know the nature of the roots of the quadratic equation without actually calculating them. Thus, if x z + x + 1 = 0, then, a = l, 6 = 1, c = l, 6 2 -4oc=-3. Therefore the roots are complex and unequal. Determine the nature of the roots of the following: 1. 4z 2 + 8z + 4 = 4. 2. 3z 2 -6z + 3 = 0. 3. x 2 - 7 x + 16 = 0. 4. 9 x 2 - 12 x - 6 = 0. 5. If c = 0, one root of the quadratic is 0, 88. If 6 = 0, the roots are equal numerically but of opposite signs. They are given by \ * Of 93. Equations of higher degree than the second may some- times be written in quadratic form. The following is such an equation, 2x 6 -5^ + 2 = 0. * This expression is called the discriminant of the quadratic. EQUATIONS 151 First consider x 5 as the unknown, then the equation is a quad- ratic. Solving by the formula of the preceding section 5 V25 - 16 . 1 x 3 = - - = 2, and -^ The original equation can now be written in the form (z 3 - 2) (tf- |) = 0. It is necessary now to solve the two equations a? - 2 = 0, x s - \ = 0. Whence, (x - #2) (x z + ^2x + ^4) = 0; From the second factor of the first equation, by the formula _^V^-4^4 (a) * = - -3- The approximate values of the radicals should be substituted and the values of x expressed in usable form. From the first factor of the same equation (6) x = #2. From the second factor of the second equation, From the first factor of the same equation, (d) x = V\. There are, altogether, six roots; these should be expressed as decimals to four figures. The solution of this problem should be carefully mastered as a typical case of higher equations in quadratic form. Use table I for evaluating radicals. 2. z + 2a; + l = 0. 3. a? -3s + 2 Vs 2 -3x + 2 = l. 4. Vs 1 - 5 -Vx 1 = -6. CHAPTER XI THE LINEAR FUNCTION AND THE STRAIGHT LINE 94. The most general function of the first degree in one variable is of the form: (1) f(x) = mx + b, where m and b are constants. The most general equation of the first degree in two variables is of the form: (2) Ax + By + C = 0, where .A, B and C are constants. Solving (2) for y t (3) y=-* x -C which is of the form (4) y = mx + 6. Equation (4) is of the same form as (1) except that y is written for / (x). Equation (3) shows that (2) can be written in the form of (4) . Since (3) is obtained from (2) by transposing and dividing by a constant, it is virtually the same equation. Since (4) is of the same form as (3), any conclusions drawn from (4) will be valid for (3) and consequently for (2). 96. Theorem. The graph of any equation of the first degree in two variables is a straight line. Using equation (4), let (xi, 7/1) be any fixed point on the locus or graph. Let (xz, y%) be any other point on the graph, chosen arbitrarily. Then by 34, yi = mxi + 6, y 2 = mxz + b. Subtracting and solving for m, 152 FAMILIES OF LINES 153 Since m is constant by hypothesis, Eq. (4), 94, and since Xz, y% is any point different from (x\, yi) it is evident the slopes of the segments joining any two points of the locus are equal. This can be true only if the locus is a straight line. If 6 is the angle be- tween the a>axis and the line, the slope of the line is FIG. 75. where (x\, yi), (x 2 , y 2 ) are any two points on the line. The angle, 6, is called the inclin- ation of the line with the axis. Note that the slope is the coefficient of x in the form (4), 94. 1. Determine the slope, intercepts and draw the lines repre- sented by the following equations: a. 2 x -{- 3y 1=0. b. x y = 14. c. x + y = 14. FIG. 76. 96. If in Equation (4), m remains fixed while b varies, there results an infinite number of equa- tions representing an in- finite number of parallel lines covering the whole plane. A set of lines having a common property is called a family of lines. If b remains fixed while m varies there results a family of lines passing through tne same point (0, 6), on the y-axis. 154 THE LINEAR FUNCTION AND THE STRAIGHT LINE When a coefficient, which is ordinarily constant in an equa- tion, is made to vary hi the manner above indicated, it is called Y / a parameter. The family of lines obtained by means of one such parameter is called a one-parameter f am- Draw several lines from each of the equations below by assigning different values to the parameter. Fir 1 77 Give & different values and draw a line for each value of 6. y = mx + 4. Give m different values and draw a line for each. 97. Converse theorem. Every straight line is repre- sented by an equation of the first degree in two variables. ' x ^lr ,,;> " i* *Z \l Af^W~ ^^ ^ T 1 ii ^| X Z~ X 1 1 1 ^s"^ f< 1- OKEj-- -H ^s^ i i 1 Y .^ O FIG. 78. Let (xi, ?/i) and fa, f/ 2 ) be any two given points on the line. Let (x, y) be any third point on the line. The slope of PI P is x x\ THEOREMS 155 The slope of PiP 2 is 3/2 -yi Since the three points are on a straight line the two slopes just written are equal. Hence y-yi = 3/2 - 3/1 x Xi Xz Xi (5) Or y-y 1= yl (x - Xl} , This equation is of the first degree in the variables x and y Hence the theorem is true. 1. Find the equation of the straight line through (2, 1) and (5, 4). Using (5) above or y-l = -f (z + 2). 7?/ + 5z + 3 = 0. 2. Find the equation of the line through (1, 3) and (2, 5). 3. The slope of a line is 2. It passes through the point (3, 7). Find its equation. Note. Use equation (5), noting that y * ~ yi = m = slope. Xz-Xi 4. The slope of a line is m. It passes through the point fa, 3/1). Find its equation. 5. Is the point (6, 4) on the line y = 4 x 2 ? 6. Are the points (3, 1), (4, 3), (6, 8) on the same straight line? 7. Find the equation of the line through (5, 7) and having a slope of . 8. Find the equation of the line through (1, 2), ( 1, 6). 9. Find the equation of the line through (3, 5) and parallel to the line 3 x - 4 y + 2 = 0. 10. What is the equation of the line parallel to the z-axis and distant 4 units from it? (Two solutions.) 156 THE LINEAR FUNCTION AND THE STRAIGHT LINE 98. Let (x, y) be the coordinates of any point P on the line AB and p = Oe the distance of the line from 0. Let the angle TOe be a. Now the projections of x and y along Oe are such that (6) x cos a + y sin a = p. FIG. 79. This relation holds for all points on AB. Therefore (6) is the equation of the line AB. Equation (6) is called the normal form of the equation of a straight line. This form is useful in prob- lems relating to the distance of a point from a line. Let A 'B' be a line parallel to AB, and let (xi, y\) be any point on A 'B'. Then if pi = Oe', (a) Xi cos a + y\ sin a = pi = p + d, where d is the distance of (x\, y\) from AB, or the distance between the lines. From (a) by transposing, (&) Xi cos a + yi sin a p = d. The left side of (&) is exactly what (6) becomes if p is transposed to the left side and (xi, yi) substituted for (x, y). Hence the distance of any point (x', y'} from a line AB may be found by reducing the equation of AB to normal form and transposing all terms to the left side, then substituting the value (x f , y') for (x, y). The result is the distance of the point (x f , y') from LINEAR FORMS 157 AB. This distance is to be considered positive if the point (zi, 7/1) and the origin are on opposite sides of the line AB, and negative if the point (zi, /i) and the origin are on the same side of the line AB, From the triangle SOT, p = b sin a = a cos a, and a/b = tan a. From 47 it is seen that a b sin a = . , cos a = , = - Va? + b 2 Va 2 + 6 2 By 94, Eq. 3, since a and b are the intercepts, if the equation of a line is Ax + % + C = 0, then a- - Va 2 -j- 6 2 It follows that -C/B -A cos a = -CIA sin a = - 'cys 2 -C 'A 2 + # 2 Thus if Ax + By + C = is the general form of the equation of a straight line, then A B -C (7) /,. . * + ~ is the normal form of the equation of the same line. It is cus- tomary to choose the sign of the radical in (7) so that the right member of the equation shall be positive. 99. The different forms of the equation of the straight line in common use may be summarized as follows: 1. Ax + By + C = 0, general form. 2. y = mx + 6, slope-intercept form. 3- y yi m(x Xi), one-point slope form (see Ex. 4, 97). 158 THE LINEAR FUNCTION AND THE STRAIGHT LINE 4. y yi = [(2/2 - y\)/(xz Zi)] (x Zi), two-point form. 5. x/a + y/b = 1, intercept form. To obtain (5), divide (1) by C and note that the values of a, 6 above are the intercepts of the line. 6. x cos a + y sin a = p or A B X -C Normal form. These forms are to be memorized. Their use is somewhat suggested by their names. In a problem, careful attention to what is given regarding a line will often suggest what form of the equation is to be used. 100. To find the distance between two points, having given their coordinates. FIG. 80. From the figure it is seen that 13, This formula is true for points in all positions. Care must be used regarding the signs of the coordinates. 101. The coordinates of a point that divides the segment joining two points in a given ratio can be found as follows: Let AB be the segment and P the point of division and r the DIVISION OF A SEGMENT 159 ratio of the two parts into which AB is to be divided. From the similar triangles in the figure AP = AM PB~ PS = r. FIG. 81. or since AM = x Xi, PS = x% x the equation becomes x x\ x = r. and In a similar way, x = y = r + l r+l If P is not between A and B the ratio r is negative. The line AB is then divided externally. 1. Find the coordinates of the points of trisection of the segment joining A(2, 1) to B(8, 4). Note. Since there are two- points of trisection the solution is to be done in two parts. First let r = |, then x = = 4 160 THE LINEAR FUNCTION AND THE STRAIGHT LINE and similarly for y. For the second point call r = 2 and use the same formula. Student complete the solution. 2. Find the middle point of the segment in Ex. 1. Note. Call r = 1 and proceed as above. 3. Find the coordinates of the point nearest (2, 1) that divides the line in Ex. 1, externally in the ratio of . Note. Callr = -f. 4. Find the coordinates of the point that divides the line in Ex. 1, externally in the ratio of 1 to 1. What is the geometric interpretation ? 102. The angle between two lines can be determined from their slopes. It is evident that in the figure = 2 0i. Hence tan $ = tan (0 2 0i) tan 2 tan 0i 1 + tan 2 tan 6\ 1 + m\m^ FIG. 82. But tan 2 and tan 0i are the slopes of A 'B' and AB respectively. The slopes are to be found by any available method and sub- stituted in the above equation. The result is the tangent of the angle <. In applying this rule it is desirable to take for 2 the larger of the two angles 2 and 0i if it is possible to determine MISCELLANEOUS EXERCISES 161 which is the larger. Then is the angle through which A B must revolve in the positive direction (counter clockwise) to bring it into parallelism with A'B'. 1. Find the angle between the lines 2x 3 i/ -f- 4 = and x y = 1. Note. Find the slopes from the equations and substitute the slopes in the above formula. 2. Find the angles of the triangles whose vertices are (1, 1); (6,8); (7, -3). Note. From the coordinates of the points in pairs find the slopes of the lines joining them and proceed as in Ex. 1. 3. What relation must hold between tan 2 and tan 0: in order that A'B' shall be parallel to AB1 4. Knowing that tan 90 = oo , show that AB will be per- pendicular to A'B' if tan 2 tan 0i = 1. 5. Show that the figure whose vertices are (0, 1); (2, 0); (5, 6) ; (3, 7) is a rectangle. MISCELLANEOUS EXERCISES 1. What is the slope of each of the following lines? (a) 10y + 3z = 6; (6) Ax + By + C = 0; (c) - = 1. 2. Find the equations of the lines satisfying the conditions given below and keep the results for later use. Draw each line. (a) Through the point (2, 1) with a slope of 2. (6) Through the point ( 3, 4) with a slope of f. (c) Through the points (1, 1) and (0, 2). (d) Through the points (-4, 1) and (3, 8). 3. Find the equation of the lino through (1, 5) and parallel to the line hi (c) above. 4. Find the equation of the line through (2, 3) and at a distance 2 from the origin. Draw the line or lines. 5. Find the equation of the line through (3, 1) and at a distance 3 from the origin. Draw the line or lines. 6. What is the normal equation of the line in (6) above? 7. What is the equation of the line whose intercepts are a = 3, 6 = 1? 8. Reduce all the equations obtained in 2 to intercept form. 9. What is the angle between the lines, 2x 3y 1=0 and x - 4 y = 3? 162 THE LINEAR FUNCTION AND THE STRAIGHT LINE 10. What is the angle between the lines x/8 y/2 = 1 and x cos 30 + y sin 30 = 4? 11. What is the distance of (1, 2) from each line in Ex. 10? 12. What is the normal equation of the line whose intercepts are a = -2, b = 81 13. The vertices of a triangle are (1, 2); (-3, 4); (4 7). Find the perimeter. Save all results. 14. What is the altitude of each ertex of the triangle in Ex. 13? 15. What is each angle of the triangle in Ex. 13? 16. What is the equation of the line through the origin and making a 45 angle with the z-axis? 17. What is the equation of the line through (3, 5) and parallel to the line 3 x - 4 y = 2? 18. What is the equation of the line through 3, 5 and perpendicular to the line in Ex. 17? 19. Find the coordinates of the points which divide ths segment join- ing (8, 4) to (3, 5) into four equal parts. 20. What are the coordinates of the point midway between (3, 5) nd the line y 2 cc + 8 = 0. 21. Show that the bisector of an angle of a triangle cuts the opposite side into segments having the same ratio as the sides adjacent the bisected angle. 22. Find the equation of the locus of all points equally distant from the points (2, 3) and (4, 1). 23. A square field has a tree near its center. Its distances from three corners are 7 rds., 8 rds. and 13 rds., respectively. Find the side of the field. 24. A regular pentagon has one vertex at the origin and one side in the x-axis. The length of a side is 24. Find the equations of the lines in which all the sides lie, respectively. CHAPTER XII 103. An equation is said to be explicit for y if it is solved for y in terms of the other variables and constants. If x and y are the variables in the equation, it is explicit for y if it is of the form y = f (*), where /(x) is any functio:: of x. The forms x 2 + y 2 = 25 and z 3 + 3:n/ + 4 = and more generally F(*,y)=0, are called implicit equations or functions in x and y. In such cases y is said to be an implicit function of x and x an implicit function of y, as is most convenient. By solving the equation for y or for x it becomes explicit. 104. Explicit quadratic function. y = ax 2 + bx + c. The graph of an equation of this form was constructed in 34. An- other example will now be discussed. Consider y = X * -f 2 X + 3. x = -4, -3, -2, -1, -0, 1, 2, 3. y = 5, 0, -3, -4, -3, 0, 5, 12. The graph is shown in Fig. 83. Answer the following ques- tions: 1. Does the curve pass through the origin? How is this determined from the equation? 2. Determine the intercepts on both axes. 3. Does any branch of the curve extend indefinitely from the origin ? That is, does the curve have infinite branches? 163 164 EQUATIONS AND THEIR GRAPHS 4. Is the curve a closed curve? 5. Is the curve symmetrical about either axis or about any other line or about the origin ? 6. What values, if any, of either variable must be excluded? That is, what values do not correspond to points on the curve? FIG. 83. These questions will now be answered with reference to the example above. Hereafter a discussion will include the answer- ing of the above questions together with pointing out any other peculiarities of the curve and the construction of the curve. 1. No, for x = 0, y = cannot satisfy an equation having a constant term. 2. At (0, 3) on the ?/-axis and at (1, 0) and (3, 0) on the re-axis. 3. Yes. For as x increases indefinitely y also increases indefinitely. 4. No. 5. Yes, about a line parallel to the y-a\is and through (-1, 0). IMPLICIT QUADRATIC FUNCTIONS 165 6. No values of x are excluded. All values of y < 4 are to be excluded. For such values of y, x is imaginary, as can be seen by substituting in the equation. What are the roots of the equation z 2 + 2 z 3 = 0? Com- pare these values with the it-intercepts. 1. Discuss y x z 10 x + 5. 2. " 105. Implicit quadratic functions. Three important cases will be considered: (a) 9 x* - 16 y 2 - 144 = 0. Solving for y, y = f Vz 2 - 16. x =- -8, -6, -5, -4, -2, 0. y = 5.2, 3.1, 2.2, 0, i, i. x = 2, 4, 5, 6, 8. y = i, 0, 2.2, 3.1, 5.2. Y FIG. 84. Here i is used to denote that the value is imaginary. The graph is given in the figure. Let the student discuss fully. This curve is called a hyperbola. (6) xy = 12. 1, 2, 3, x = I) 4, 6, 12. y= 48, 24, 12, 6, 4, 3, 2, 1. * = - i - i - 1, -2, -3, -4, -6, -12, y = -48, -24, -12, -6, -4, -3, -2, -1. 166 EQUATIONS AND THEIR GRAPHS The graph is given in the figure. Let the student discuss fully. This curve is also a hyperbola: (c) 9z 2 +16?/ 2 = 144. z=-5, -4, -3, -2, -1, 0, 1, 2, 3,4, 5. y= i, 0, 1.9, 2.6, 3, 3, .3, 2.6, 1.9, 0, i. Y Y FIG. 86. The graph is shown in the figure. Let the student discuss fully. This curve is called an ellipse. 106. The curves of 104, 105, 34, are curves of the second degree. They are called conic sections. The student should learn to recognize these curves and their equations. RATIONAL FRACTIONAL FUNCTION 167 107. Functions of the third degree will be illustrated by y = 2 x 3 5 x 2 + x + 2. 1, f, 2, 3. 0, -i, 0, 7. y = -3, 0, 1, f, FIG. 87. * The graph is given in the figure. It is typical of all explicit cubic functions. Let the student discuss fully. 108. Rational fractional function. Consider the equation 1 1 2z~ z (z - 1) (z - 2) The second form is more convenient for calculating: z^-oo, -2, -1, -J, 0, }, 1, 1, f, 2,3, 4,oo. y= o, -,v, -I, -A, TOO, 3^-, f 00, -|, TOO, i, ^ T , o. The graph is shown in the figure. It should be noted that the infinite values of y occur at the zero values of the denominator. At these values the definition of continuity does not hold. The function is said to be discontinuous at such points. The value of x which gives an infinite value of y is called an infinity of the 168 EQUATIONS AND THEIR GRAPHS function or a pole of the function. Let the student discuss the example fully. Y X- Y' FIG. 88. 109. As an example of irrational functions, consider y = Vx? - 5 x 2 + 6 x = Vx (x 2) (as 3). FIG. 89 EQUIVALENCE OF EQUATIONS 169 The second form is most convenient for calculating: x = Q, 1, i 2, I, 3, 4, 5._ y = 0, V2, V|, 0, i, 0, V8, V30. The curve is shown in the figure. Let the student discuss fully. Construct the graphs and discuss fully the following: 1. y = x*-x~ 1 + 5. 7. 7/2 = TC J 2. xy + y* = 23. 8. xy + y 2 = 0. 3. y = x$. 9. z - 1 = ! 4. y = V2 x 3 + x" - 2 x + 2. 10. y = (x - 3) 3 . g (1 yZ\ y y- _|_ g ]_]^ ^ = 6. y = [ 110. Simultaneous equations of the second and higher degrees in two unknowns can be solved (at least approximately) by means of their graphs. To do this construct the graphs of both equations on the same axes and to the same scale. The measured coordinates of the points of intersection of the graphs will be the solution of the pair of equations. The slide rule and tables of squares and cubes should be employed to facilitate calculation. 1. y + x z = 7 and y 2 + x = 11, find x and y. 2. x z + y 2 = 25 and x + y + 1 =0, find x and y. 3. x z + y- = 25 and z 2 /9 + 2/ 2 /36 = 1, find x and y. 4. z 2 /9 + 2/Y36 = 1 and z 2 /4 - y*/lQ = 1, find x and y. 5. y = x 3 and x z + y 2 = 25, find x and y. 6. x 2 + i/ 2 = 9 and i/ = sin x, find Z and ?/. 7. y = cos x and ?/ = sin x, find z and y. 111. The question of equivalence of equations and systems of equations will not be discussed systematically. A few examples illustrating the meaning of the term and impressing the need of care in checking of results will be given. 170 EQUATIONS AND THEIR GRAPHS Two equations in the same unknown are equivalent when every root of each is a root of the other. Thus x 2 - 3 x - 4 = and 5 z 2 - 15 x - 20 = are equivalent. Let the student solve and verify the state- ment. The equations are not equivalent. Let student solve and verify. The follow- ing operations on an equation lead in general to an equivalent equation. 1. Multiplication or division of both members by the same known number. 2. Addition or subtraction of the same expression on both sides of the equation. 3. Clearing of fractions in most ordinary cases. The following operations may not lead to an equivalent equation. 4. Multiplication or division of both members by an expres- sion containing the unknown (except as noted in (3)). 5. Clearing an equation of radicals. 1. Solve Vx-\- 5 Vx 4 = 9; then solve Vx + 5 + Vx 4 = 9 and test results by substitution in the original equations. 2. Are x 7 Vx 5 = and x z 15 x + 54 = equiv- alent? A system of simultaneous equations is equivalent to another system if every solution of each system is a solution of the other system. In solving systems of equations it is best to substitute all results hi the original equations. 1. Show, by solving and substituting, that the system x y = 1 . $ x y = 1 is equivalent to * EQUATIONS 171 2. Determine whether the systems 9z-6?/ + 12 = are equivalent. 112. Some cases of systems of quadratic equations in two unknowns are easily solved. A few commonly occurring cases will be given below: (a) One equation linear and one quadratic. This case was given in Chap. I, 10. (6) Both equations of the second degree and containing only the squares of the unknowns. Thus ax 2 + by 2 = c, a'x 2 + b'y 2 = c'. First, regard these equations as linear in which x 2 , y z t instead of x, y, are the unknowns. Solve for the values of x 2 , y 2 . Take the square root of each value of x 2 , y 2 for the values of x, y. (c) All terms containing unknowns are of the second degree but not necessarily the squares of the unknowns. Thus x 2 + xy = 10, y 2 - xy = 12. Substitute y = mx in both equations and get x 2 + mx 2 = 10, m 2 x 2 mx 2 = 12. Solving each of the last equations for x 2 and equating results, 2= 10 12 m + 1 m 2 m Solving the last equation for m, m = 2.652 and m = -0.452. Whence by the equation y = mx, there is obtained y = 2.652z and y = -0.452 x. 172 EQUATIONS AND THEIR GRAPHS Substituting the values of m in x 2 = ; in succession, m + 1 x 2 = 2.738 whence x = 1.652. Substituting the value of x in y = 2.652 x gives y = 4.381. Similarly using the other value of m will give other values of x, y. These should be tested by substituting in the original equations. x 2 + y z = 25 (d) To solve the system , ( oxy = 66 Divide the second equation by 3 and add to the first equation, obtaining Taking the square root x -f- y = 6 (two equations). By subtracting instead of adding as above there is obtained x z 2 xy + y z = 14. Taking the square root x y = Vl4 = 3.74+ (two equations). Solving now these four equations of the first degree in pairs will give the desired solution. Test the results in the original equations. (e) The system (rf + ff-8 i * + y =2 can be solved. First divide the first equation by the second, member by member, x* xy + ?y 2 = 4. This equation together with the second equation form a system like the one described hi (a). This system has, therefore, been treated in Chapter I. Complete the solution and test results. EQUATIONS 173 i 2 4- if = 25 1. Solve the system j 4a;2 + 6?/2 = ^ ( -r 2 -J- ifl = 2. Solve the system 3. Solve the system 4. Solve the system 5. Solve the system 6. A piece of cloth on being wet shrinks 10 per cent in length and 6 per cent in width. The total loss of area is 5 sq. yd. How many square yards were in the piece originally? 7. If $1000 at a certain rate for a certain time at simple interest amounts to $1250 (principal and interest) and if the same principal for 3 years less time at 2 per cent lower rate amounts to $1200, find the rate and the tune. 8. A pole stands on a tower. A man 5' high (to his eye) standing on level ground finds that at a certain distance from the foot of the tower the angle subtended (at his eye) by the tower is the same as that subtended by the pole. The tower is 50' high and the pole 80' high. Find the distance of the man from the foot of the tower. 9. A garden plot adjacent a wall is to be fenced. The area is to be 160 sq. yds. The length is to the breadth as 3, 2. Find the dimensions of the plot. (Fence on three sides.) Two solutions. 10. The sum of the squares of two numbers is 83. Their difference is 4. Find the numbers. 11. The area of a circular race track is 150,000 sq. ft. The inner diameter is to the outer diameter as 14 to 15. Find the inner and outer diameters of the track. 12. A rectangle has an area of 135 sq. rds. If lines are drawn from two opposite vertices to the diagonal joining the other vertices divide that diagonal into three equal parts. Find the dimensions of the rectangle. CHAPTER XIII P=P' TRANSFORMATION OF COORDINATES 113. It is of great advantage, in certain problems, to be able to simplify or to change the form of an equation. Certain troublesome terms may be removed or some other change may be made. One common way of attaining these results is to substitute for the variables certain linear functions of new variables. This is called a linear transformation. Such a transformation has the effect of moving the axes to a new posi- tion with reference to the curve whose equation is thus trans-, formed without affecting the fundamental properties of the curve or the degree of the equation. (a) To rotate the axes of coor- dinates through an assigned angle, 6, without moving the origin. The equations of transformation can be determined by considering only one point. For all points will be sim- ilarly affected by the transforma- tion. Consider the point P(x, y) referred to the axes OX, OY. Let the coordinates of the same point be x', y' referred to the new axes, OX', OY'. In the new posi- tion call P = P'(x', y'} which, of course, is the same point as P(x, y} referred to the old axes. From the figure x = Om, y = mP, x' = 01, y' = IP, Hence x = Om = On + nm. But On = x' cos 6 and nm = y' sin 0. Therefore (1) x = x' cos 6 y' sin 6. 174 y FIG. 90. LINEAR TRANSFORMATION 175 In a similar manner, (2) y = x' sin 6 + y' cos 6. Equations (1) and (2) are the ones desired. When 6 is given x and y are expressed as linear functions of x' and y'. "When 6 is not known it may be found when certain other conditions are given. (6) To move the origin without changing the direction of the axes. It is easily seen from the figure that Y' FIG. 91. (3) x = x' + h and (4) y = y' -\- k, where the new origin is the point O'(h, k), referred to the old axes. 1. By use of (3), (4), move the origin to the point (2, 3) in the equation x z + y 2 - 4 x - 6 y = 12. Substituting x = x' + 2, y = y' + 2, into this equation, (x r + 2) 2 + (y f + 3) 2 - 4 (x' + 2) - 6 (y r + 3) = 12. Expanding and collecting, this reduces to z' 2 + y' 2 = 1. The primes may now be dropped if we remember that this equation is to be referred to the new axes. Hence the last equation may be written : x 2 + y* = 1. 2. Using (1), (2) rotate the axes 45 in the positive direction in the equation xy = 12. Substituting the values of x, y from (1), (2), for 6 = 45, into this equation (x' cos 45 + y' sin 45) (x' sin 45 - y' cos 45) = 12, 176 TRANSFORMATION OF COORDINATES or (0.707 x' + 0.707 y'} (0.707 x' - 0.707 y') = 12, or 0.5z' 2 -0.5*/' 2 = 12, or dropping primes, a? - y* = 24. This equation represents the same curve referred to the new axes. 3. Determine h, k so that the first power terms in x and y shall disappear from the equation z 2 4 1/ 2 2x -\- &y = 1. Note. After substituting from equations (3), (4), collect the coefficients of the first power of x and equate the result to 0. Similarly equate the coefficient of y to 0. The resulting two equations will determine h and k. 4. Determine 6 so that the xy term shall disappear from x z + xy + 2 y* + x = 0. Note. Use Equations (1) and (2) and proceed as in the last exercise, putting the coefficient of xy equal to zero and solv- ing for 6. 5. In y = mx + &, rotate the axes 90 in the positive direc- tion. What is the meaning of m in the new equation? Of 6? 6. In ^ + |T = 1, rotate the axes 30. Note the form of AO y the resulting equation. 7. Using the first of equations (3) determine h so that the term in x z shall disappear. Draw graph of original and of new equation 8. By any method above remove the xy term from x z - f + 2 xy + 3 = 0. X 2 2/ 2 9. Change TZ + q = 1 * polar coordinates, using the transformation x = r cos 0, y = r sin 6 (73, Ex. 1J, 12). 10. Change y* = 8 x to polar coordinates. CHAPTER XIV CONIC SECTIONS 114. The conic sections constitute the geometric aspect of integral functions of the second degree. In these are included all integral equations in two variables where at least one term is of the second degree. For this reason the conic sections are called loci of the second order, curves of the second order, or curves of the second degree. Historically the geometric idea was developed first. The correlation of the geometric with the algebraic (analytic) idea with respect to these curves has been of much value in the study of the laws of nature. Definition. The locus of a point which moves in a plane so that the ratio of its distances from a fixed point and a fixed straight line is constant is called a conic section * or for short a conic. This definition, based on the discovery of the property of these curves by the Greeks, furnishes a convenient starting point for an introductory analytic study of the curves. The fixed point referred to in the definition is called the focus of the conic, and the fixed line the directrix of the conic. The conies are conveniently classified, for purposes of ele- mentary study, according to the different values which the ratio, referred to in the definition, may take. This ratio is called the eccentricity of the conic and will be denoted by e. 115. Ratio equal to one, e = 1. Parabola. Let F be the focus, DD' the directrix and P (x, y) any point on the curve. * For proof that the curves of intersection of planes with a right circular cone have this property, see Wentworth Geom., Rev. Ed., p. 458, or some treatise on conic sections. 177 178 CONIC SECTIONS We wish to find the equation of this curve. Choose the origin on the curve midway between the focus and the directrix. From the hypothesis and the figure we can write FP = KP, OF = MO = p. Expressed in terms of x, y and p the relation FP = KP becomes FIG. 92. or (1) ?/ 2 = 4 px. This is the desired equation of the parabola in the standard form. Let the student discuss the curve. This is the same kind of curve as the one given in 104, the difference being the position with reference to the axes. 1. Find the total distance across the curve through the focus perpendicular to the aj-axis. This double ordinate through the focus is called the latus rectum. 2. Move the origin to the point (h, k) and note the change in the equation. 3. Rotate the axes 90 in the negative direction and note the form of the resulting equation. 4. Find the equation of a parabola (standard form) that passes through (5, 2). Determine the distance from the focus to the directrix and the distance from the focus to the origin-. 5. Transform equation (1) so the origin will be at F. 6. Transform the resulting equation of example 5, to polar coordinates with pole at focus and polar axis the z-axis. Note form of equation. 7. Show that for any two points on a parabola, the squares of the ordinates are proportional to the abscissas (equation to be in standard form). Give a geometric interpretation of this theorem, that will be independent of the position of the axes or the form of the equation used. ELLIPSE 179 8. Simplify the equation y = 4 z 2 6 x + 40, by putting it in the form y = kx 2 . Note. The constant and the term in x may be removed by use of equation (3), (4), 113. 9. Simplify and discuss 4 ?/ 2 6 + 3 = x 5. 10. Derive the equation of a parabola passing through (1, 1), (3, 5), (0, 0). Assume equation of form (y a) 2 = k (x I) and determine a, k, I and write the equation accordingly. 116. Ratio less than one, e < 1. Ellipse. Use the nota- tion of 115 except that MF = p = MA + AF, MA ^ AF. Now from the figure and the hypothesis we can write AF - _ ^1 ~ ~ MA ~ KP AF = e-MA. Solving for MA, AF, MA = AF = 1+6 ep r+v Now FP = e KP for all points of the curve. Expressing this relation in terms of the coordinates of P (x, y) there is obtained 180 CONIC SECTIONS which reduces by clearing of radicals to (1) (1 - e 2 ) z 2 - 2 epx + y z = 0. This is the equation of the ellipse in the position shown in the figure. 1. Discuss the curve from the above equation. 2. Move the origin to the point [_ a , j to remove the term in the first power of x. (See 113 (6).) CT) 3. Call 1 g = a and (1 e 2 ) a 2 = b 2 in the equation ob- J. > tained in example (2) and show that the equation reduces to (2) -2 + ^ =1 - a 2 o 2 This is the standard form of the equation of the ellipse. The origin is now 0' and the y-axis is O'Y' in the figure above. 4. Note the meaning of a, b in the figure and show that the intercepts on the x-axis are a and +a, and on the ?/-axis the intercepts are 6 and +6. The values a and 6 are the semi- axes of the ellipse. The value a is the semimajor axis and the value 6 is the semiminor axis; 0' is the center. 5. Discuss the curve by use of the equation (2) above. a * 6- Solve this equation for p in terms of a and e. The result is a (1 - e 2 ) p = i - '- = MF. e 7. Find the value of MA in terms of a and e. 8. If c ae, show by use of the values of a and 6 above that a 2 6 2 = c 2 for any ellipse. 9. Show by use of the result of Ex. (8) that if e = the ellipse becomes a circle. Note. This supposition makes a = b. 10. Find the equation of the locus of a point that is always 16 units from the point (3, 1). HYPERBOLA 181 11. A circle has its center at (3, 2) and passes through the point (8, 11). Find its equation. Note. Formulate the distance between (3, 2) and any point (x, y) of the curve and equate this expression to the given radius. Free the equation of radicals. 12. Find the equation of the ellipse, in standard form, if the eccentricity, e, is and the curve passes through (3, 4). Note. Form two equations and determine a and b of equa- tion (2). 13. Find a, b, c, e and p in the ellipse 25 x* + 144 y 2 = 1500. 14. Prove that for any ellipse FP + F'P = 2 a where P is any point on the curve. 15. What is the standard equation of /v2 ^y2 16. With the standard equation -^ + j- = 1, move the origin y ^i to the left-hand focus ( c, 0), then change to polar coordinates. Note the form. Compare with Ex. (6), 115. 17. Find the length of the double ordinate through one focus x 2 7/ 2 in -5 + r; = 1, and find the distance of the end of this ordinate a 2 o 2 from 0, and from the other focus. See Ex. 1, 115. 117. Ratio greater than one, e > 1. Hyperbola. With the same notation as in the last section and from the annexed figure may be written (1) (e 2 - 1) z 2 + 2 epx - y* = 0, e > 1, where the origin is at A and where M A + AF = p. Since (1 e 2 ) < 0, move the origin to ( _ , 01. The above equation becomes or a? - 1 - e 2 (e 2 - I) 2 182 CONIC SECTIONS Calling e z _ i = a> and a 2 (e 2 1) = 6 2 , the last equation may be put hi the form A A (2) .K_. Mj -p > e- Da a-->*-a- D FIG. 94. This is the standard form of the equation of the hyperbola referred to 0' is the origin and O'Y' the y-axis. 1. If c = ae, show that a 2 + fr 2 = c 2 . 2. Discuss the curve from equation (2). /y2 qj2 3. Discuss the curve whose equation is -5 ^ = 1, and a 2 b z compare with the curve of equation (2). The curve of this exercise is called the conjugate hyperbola of the curve of equation (2). Note the intercepts of both curves on the axes. x y x if 4. Discuss the two curves ^ 77 = 1 and T^ ?r = lo 9 lo 9 1. Draw the curves on the same axes. 5. Prove that for any hyperbola F'P FP = 2 a, equation (2) . 6. With equation (2) rotate the axes 45 in the negative direction. Compare the result with the example of 105 (6). x 2 ii 2 7. With ZT = 1, move the origin to (c, 0) and change to DIAMETER 183 polar coordinates. Note the form and compare with the polar forms of the parabola and the ellipse previously derived. 4) ft /^ ni, 8. Find e, a, b, c, p for the curve ^= ^ = 1. AD ID 9. Find the equation, in standard form, of a hyperbola of eccentricity, e = 2, and passing through (7, 5). Find the equation of the conjugate hyperbola. 10. Find the equation of a hyperbola in standard form which passes through (3, 5) and (5, 7). 118. Diameter. A line which bisects a system of parallel chords of any curve is a diameter of the curve. All the conies have diameters. To illustrate, consider the ellipse whose equation is ^j.^!_i a 2 ' 6 2 Let P' (x', y') be the midpoint of any chord, AB. Let P (x, y) be a variable point on AB. Call P'P = r, and consider r positive if P is between P' and A, negative when P is between P' and B. Now from the figure write = sin0. FIG. 95. If P moves to A or to B its coordinates must satisfy the equation of the ellipse. Solving the last two equations for x } y and y? ip substituting in z + -^ = 1 gives L 2rx / cos0 ?/ 2 + 2 ry' sin e + r 2 sin 2 _ 184 CONIC SECTIONS When P is on the curve the roots of the last equation with regard to r are equal but of opposite signs. Hence the coeffi- cient of the first power of r must be 0.* Therefore, x' cos & y' sin 6 _ _ ~*~ ~~V~ 6 2 or y' = j~ x' cot 6. This is a linear relation between x f , y', the coordinates of the midpoint of any (consequently all) chords making 'a given angle, 6, with the axis. The equation, therefore, holds for all points of the bisector of a parallel system of chords, and is the equation of the diameter which bisects these chords, that is of CD. x z y z 1. Find the equation of the diameter of an ellipse =^7 + % = 1, ID y which bisects the system of chords which make a 30 angle with the re-axis. Note. Adapt the method above to this case and work out in full. 2. Find by a method similar to the above, the equation of the diameter of the parabola which bisects all chords which make a 45 angle with the z-axis. 3. Prove that if the slope of a diameter of the ellipse, + -i a 2 + 6- ~ L| is m and the slope of the corresponding system of chords is m', 6 2 then m,' m' = -= a 2 4. Show that if the diameter of an ellipse has a slope m, and the slope of the bisected chords is m', then the diameter which bisects the chords parallel to the first diameter has m' for its slope. The two diameters referred to in this exercise are called conjugate diameters. * See 92, Ex. 5. ECCENTRIC ANGLES 185 x ii z 5. Find the equation of diameter of - + ^ = 1, which passes y xo through the point (1, 1) and find the equation of the diameter conjugate to this. x z y-i 6. Find the equation of the diameter of ^ ^ = 1, which bisects the chords which are parallel to 2 x + 3 y = 0. 119. Eccentric angles, (a) Ellipse. In the figure the large circle is called the major auxiliary circle, the small circle the minor auxiliary circle of the ellipse. P', P are points on the ellipse and major circle having the same abscissa. The angle, 6 = A OP', is called the eccentric angle of the point, P, of the ellipse. FIG. 96. From - z + f^ = 1 and x' 2 + y' 2 = a 2 , it is evident that for \Jv t/ x = x' the corresponding values of the t/'s are related by x y 1. Suppose the ellipse -5 + r^ = 1 be cut into narrow strips CL parallel to the y-axis (approximately rectangles). From the x* y* relation above show that the area of the ellipse -| + p = 1 is 186 CONIC SECTIONS b/a times the area of the circle x 2 + y* = a 2 . That is, that the area of the ellipse is nab. 2. Show that the area of the ellipse is Vl e 2 times the area of the circle given in example (1). 3. Show that if <j> is the angle between the plane of the circle z 2 + y z = a 2 and the plane of the X Z 11" ellipse -5 + rj = 1, then sin <j> = e FIG. 97. and cos <j> = Vl e 2 and that con- sequently from the results above cos <j> = b/a = \/l e 2 . (Remember e = c/a, c 2 = a 2 & 2 .) 4. Find the equation of the ellipse which is the projection of the circle x z + y z = 25 on a plane at an angle 30 with the plane of the circle. (6) Hyperbola. In the figure P, P' are corresponding x 2 y* points on the hyperbola -5 jj^ = 1 and its auxiliary circle 2 + y* = a 2 - The angle is called the eccentric angle of the point P, on the hyperbola. It is easily seen that x = a sec 0, Y FIG. 98. These relations are closely connected with a very interesting set of functions, known as the hyperbolic functions. These GENERAL EQUATION 187 functions have properties quite similar to those treated in Chapter VIII. (See McMahon Hyperbolic Functions.) 120. The most general equation of the second degree hi two variables is of the form (1) Ax* + Bxy + Cy 2 + Dx + Ey + F = 0. Under certain conditions the terms of the first degree may be removed by moving the origin. Assume x = x' + h and y = y' + k and substitute in (1) ; the result is (2) Axf + Bx'y' + Cy'* + (2 Ah + Bk + Z>) x' + Ek + F = 0. The terms of the first degree can be removed if and only if simultaneously the equations, 2Ah + Bk + D = 0, Bh + 2 Ck + E = 0, hold for finite values of h and k. Solving for h and k gives 2CD-BD , 2AE -BD ~~ ~~ These values are finite if B 2 4 AC = A ^ 0. Therefore, the first degree terms can be removed by moving the origin if A 7* 0. The term in xy may always be removed by rotating the axes. It will then be sufficient to discuss in detail the equation, (3) Ax 2 + Cy z + Dx + Ey + F = 0. I. Consider this equation when A ^ 0, C ^ 0. Now A 5^ 0. The terms of first degree can be removed. The resulting equa- tion will be of the form Ax 2 + Ci/ 2 + F' = 0. (a) If A and C are like signed and F' 7* 0, the equation represents an ellipse, real if the sign of F' is opposite that of A and C, imaginary if the sign of F' is the same as the sign of A and C. If F' is zero, the ellipse is a point ellipse. 188 CONIC SECTIONS (6) If A and C are unlike signed and F' ^ the equation represents a hyperbola. If F' = 0, the left member of (3) breaks up into two lineal* factors and represents two straight lines through the origin. II. Consider the case where either A or C vanishes, say A = and C ^ 0. Now equation (2) becomes (4) Cy + Dx + Ey + F = 0. If C = and A ^ 0, the equation reduces to (5) Ax* + Dx + Ey + F = 0. By moving the origin to a point (h, 0) equation (4) may be reduced to (6) Cy z + Dx = 0, and by moving the origin to (0, k) equation (5) may be re- duced to (7) Ax* + Ey = 0. Equations (6), (7) represent parabolas. In case the equations (4), (5) take the form (8) Cy 2 + F' = 0, (9) Ax z + F = 0, respectively, they represent pairs of straight lines parallel to the axes. 121. Confocal conies. Two conies are confocal when they have the same foci. Consider the two conies whose equations are: The foci of these conies will coincide if o 2 6 2 = a\ 2 2>i 2 , or a* - ai 2 = 6 2 - &! 2 . Let ai 2 = o 2 + X and 6i 2 = 6 2 + X. Sub- stituting in (2) : 2 CENTERS OF CONICS 189 The curve (3) is confocal with the curve (1) for all values of X. Why? /*2 nj'Z 1. Find the equation of a conic confocal with ^ + ^ = 1 AO \j and passing through (5, 6). x z y 2 Note. Write ; 7 r + n , . = 1. Substitute the coordi- ib + A y + x nates of the given point hi this equation and solve for X. Having found X, substitute its value in the above equation. 2. For what values of X will the equation 16 + X 9 + X represent an ellipse? For what values of X will it represent a hyperbola? See 116-117 and determine what values of X make this equation like those referred to. x 2 y 2 3. Find the equation of a conic confocal with ^ = 1 *7 J.D and passing through (5, 7). 4. Determine whether 8 x 2 + 12 y 2 = 96 and 3 x 2 - 8 y 2 = 24 are confocal conies. 122. Centers of conies. The ellipse and hyperbola haver two perpendicular axes of symmetry and therefore have a center of symmetry. This point is called the center of the conic. The circle is considered as a special case of the ellipse. When the equations of the ellipse and hyperbola are in standard form the centers are at the origin. When the equations are not in standard form the center, referred to the original axes, is found by the values of h, k by which the terms of first degree are removed. The values of h and k being the coordinates of the center referred to the original coordinate axes. It is to be understood that the xy term has first been removed from the equation before applying this method. Full treatment of the method of finding the centers of conies cannot be entered into in this course. Some examples will be of use in showing how to determine the center of a conic in certain cases. 190 come SECTIONS (a) Find the center of the circle x 2 + y* Qx + 8y = 0. This equation may be written in the form x 2 - 6 x + 9 + y 2 - + 8 y + 1 6 = 25, or (x - 3) 2 + (y + 4) 2 = 25. It is evident that if the origin be moved to the point (3, 4) that the first degree terms will disappear and the equation would be in standard form. Therefore, the center of the circle is the point (3, 4), referred to original axes of coordinates. Let the student draw the curve and verify this result. (6) Find the center of 4 x 2 + 6 y 2 + 12 x - 24 y = 3. This equation can be written in the form 4x 2 + I2x + 9 + 6?/ 2 - 24 ?/ + 24 = 36, or (2x + 3) 2 + 6 (y - 2) 2 = 36. It is evident if the origin be moved to ( f, 2) the first power terms will vanish and the center will be at that point. 1. Solve each of the above problems by the method of mov- ing the origin, 113 (6). 2. Show that center of the circle x z + y z + Dx + Ey + F = is at the point ("o i ~"S~~)' 3. Show that the equation of the circle whose center is (a, 6) and radius r is (x a) 2 + (y 6) 2 = r 2 . 4. Find the center of x 2 - y 2 - 3 y + 24 = 0. 5. Find the center of x z + y 2 - Sx + IQy - 20 = 0. MISCELLANEOUS PROBLEMS 1. Reduce 2 x* - 5 xy - 3 y z + 9 x - 13 y + 10 = to one of the standard forms and determine a, b, c, e, p as the case may require. 2. What is the equation of the circle of radius 10 and center at the point (2, 3). (See Ex. 3, 122.) 3. What is the length of the common chords of x 2 + j/ 2 = 8 and 9 x 2 + 4 y* = 36? (The common chord joins two points of intersection.) 4. Find the equations of the straight lines which coincide with the common chords in Ex. 3. 5. Find the standard equation of the hyperbola whose foci are the points (3, 0) and (3, 0) and eccentricity 1.5. MISCELLANEOUS PROBLEMS 191 6. See if the curve ^ = 2 x VT/Q touches the curve 3 z 2 6 if = 1. Note. Solve simultaneously and determine whether the curves cut or just touch each other. 7. Find the equation in standard form of an ellipse whose foci are ( 3, 0) and (3, 0) and eccentricity 2/3. 8. Find the equation of the run of a 6" stove pipe cut at an angle of 30 with the axis of the pipe. (Standard form.) 9. Find the equation of the boundary of the shadow of a circle of radius, r, on a plane making a 45 angle with the plane of the circle. (Standard form.) (Light vertically above plane.) 10. Find the equation of a circ e passing through the points (1, 2), (3, 5) and (-1, 4) Find the lengths of the three chords joining these points. Note. Use (x a) 2 + (y 6) 2 = r 2 and deter- mine a, 6 and r. 11. Find the equation of a parabola, in standard form, which has its focus at (4, 0). 12. If x = vt and y = aP, eliminate t and discuss the locus of the resulting equation. 13. Prove that two elliptical gear wheels of the same size and shape will work together smoothly if connected by rods joining their foci as shown in the figure. 14. Determine the kind of conic represented by the following: (a) 4x 2 + y z - 13z + 7y - 1 = 0. (6) 3z 2 -4j/ 2 - 6^ + 9 = 0. (c) 7 z 2 - 17 xy + 6 y 2 + 23 x - 2 y - 20 = 0. 15. Find a, 6, c, e, p and the coordinates of the center of: 81 =0. (a) (6) 5z 2 (c) 5x 2 5xy - 7y 2 - 165x = 0. 1320 = 0. 16. Find the equation of a conic confocal with Tc + q = 1 ^d passing through the point (a), (6, 5); (6), (3, 2). 17. In the figure AB and CD are the axes of the ellipse. RT = % AB, RS = % CD. Show that if T be made to move on CD while S moves on AB, then R traces the ellipse. (This is the principle of the ellipsograph.) 18. By use of the property of the curve expressed in Ex. 14, 116, devise a method of constructing the curve by the intersections of pairs of arcs whose centers are at the foci. 19. By use of the definition of the parabola, 116, show how to construct 192 CONIC SECTIONS a parabola by intersection of arcs whose centers are at the focus with straight lines parallel to the directrix. 20. By use of the property of the curve expressed in Ex. 5, 117, devise a method of constructing points on a hyperbola by intersections of arcs whose centers are at the foci. 21. Show that an ellipse can be drawn by using a string of length 2 a fastening the ends of the string at the foci and holding a pencil against the string while drawing the curve. See figure below. JL-- Pencil point Ex. 21. are the 22. Show that the points (2, 2), (-2, -2) and (2 V3, -2 vertices of an equilateral triangle. 23. Prove that (10, 0), (5, 5), (5, -5), (-5, 5) are the vertices of a rapezoid. 24. In any triangle show that a line joining the middle points of two rides is parallel to the third side and equal to half of it. Note. Use two-point form and compare slopes. 25. Find the equation of a circle whose center is at (3, 2) and its radius 4. Discuss the curve y = x 2 2 x 3. Find the point of intersection of x 7 y = 25 with x z + y 2 = 25. 26. 27. Plot. 28. 29. tf- Change x 2 -f y z 25 to polar coordinates. Move the origin so as to remove the first degree terms from a? 1 12 = and discuss the curve with reference to the new axes. CHAPTER XV THEOREMS ON LIMITS, DERIVATIVES AND THEIR APPLICATIONS 123. Theorem I.* The limit of the sum of two or more f infinitesimals, 37, is zero. Let 81, 5 2 , 5 3 be three infinitesimals. By definition each must become and remain less than some arbitrarily small number, | say, where e is arbitrary. Hence at some stage and o thereafter, e e 8t< , 8 2 <3, 8 3 < 3 and 81 ~\~ 82 -\~ 83 <C 3 ^ = e. Since c is arbitrary the sum 5i + 5 2 + 5 3 satisfies the definition of a limit and we conclude (see 37) Lim (Si + 5 2 + 8 3 ) = 0. 124. Theorem H. The limit of the sum of two or more variables is the sum of their limits. Let x, y, z be three variables and a, 6, c their respective limits. Let x a = Si, y b = 5 2 , z c = 5 3 .f Now from the definition of limit, Si, S 2 , 8 3 , each has zero for its limit (see 37d). Write x = a + Si, y = b + 5 2 , z = c + 5 2 * This and the three following theorems may be treated as assumptions if preferred and proofs omitted. t In this and following theorems the term " more " is not to be interpreted to mean an infinite number. J If 2 < c, 83 will be negative and similarly for the others. 193 194 LIMITS AND DERIVATIVES and add x + y + z = a + 6 + c + 61 + 8 2 + 5 3 . Ihe limit of the right side of this equation is a + 6 + c (Theo- rem I). Therefore, Lira (x + y + z) = lim x + lim T/ + lim z. 125. Theorem HI. The limit of the product of two or more variables is the product of their limits. With the same notation as in 124, write Lim xyz = Lim (a + 61) (6 + &) (c + 5 3 ) = Lim (a&c + a8z8 3 + b8 t 8 3 + cdi8 3 + ab8 s +0,082 + bcSj + 818283) = abc, since every term on the right after the first has zero for its limit. In a similar way the proof can be extended to any number of variables. The theorem is, therefore, true. 126. Theorem IV. The limit of the quotient of one variable divided by another is the quotient of their respective limits. With the same notation as above write r . x T . a 81 Lim - = Lim r - y b - 8 2 By long division the right member may be written so that since the numerator of the second fraction has zero for its limit. Therefore by 42, III, the fraction has zero for its limit. 127. Definition and formation of the derivative of a func- tion. All numbers may be thought of as values which a variable may assume. Any number may be considered as the sum of two numbers, one a value which a variable may assume at some instant, the other an increment to the first so that the sum is a subsequent value of the variable. This way of think- mg of numbers and in particular of symbols or variables which DERIVATIVE 195 assume number values yields a very useful instrument for solving certain kinds of problems which we have hitherto been unable to attack. In order to make use of this idea it is necessary to learn to calculate and to manipulate a function called the derivative of another function. The derivative of a function may be defined as the limit of the ratio of the increment of the function to the increment of the independent variable when the increment of the variable has zero for its limit. In symbols this definition may be formulated as follows. Let the function be V = /(*) Then Lim if it has a limit, is the derivative of the function /(x) with respect to x. The steps in calculating the derivative are: *-/(*). (1) 2/ + Ai/=/(x + Ax). (2) Subtracting, A?/ = f(x + Ax) -/(x). /<>\ T^- -j- u A ty f( x + Ax) f(x) (3) Dividing by Ax, ^| = - ^ (4) Taking the limit, &- Urn - Eh. *' The symbols -p, Lim rr*, /'(x), D x y, -r- f(x) are to be used UX Ax ()iA.C ('.(' synonymously in this course as indicating the derivative of y with respect to x, where y = /(x), or as the derivative of /(x) with respect to x, where y = /(x). 1. Calculate the derivative of /(x) = x 3 + 2 x + 1 with respect to x. Write 196 LIMITS AND DERIVATIVES Giving x, y increments, Ax, Ay respectively y + Ay = (re + Ax) 3 + 2(x + Ax) + 1 = x 3 + 3 x 2 Ax + 3 x(Ax) 2 + (Ax) 3 + 2 x + 2 Ax + 1. Subtracting, Ay = 3 x 2 Ax + 3 xAx 2 + Ax 3 + 2 Ax. Dividing by Ax, ^ = 3 x 2 + 3 xAx + (Ax) 3 + 2. Taking the limit, = 3 x 2 + 2, since all other terms become zero when Ax 0, 124. 2. Calcula to x. Write 2 - 2. Calculate the derivative of /(x) = re .. , , with respect X ^ M/ 2x = x - 1+x A 2 (x + Ax) A 2 Ax = Ax 77; Ax (1 + x + Ax) (1 + x) dx (1 + x) 2 3. Calculate the derivative with respect to x of y = 3x 3 -5x 2 + 2x-l. 4. Calculate the derivative with respect to x of y = 2 - x + 2 x 2 - 3 x 3 . 5. Calculate the derivative with respect to x of 6. Calculate the derivative with respect to x of 1 DERIVATIVES 197 7. Calculate the derivative with respect to x of 8. Calculate the derivative with respect to x of y=(x- 3)4. Note. Expand first. 9. Calculate the derivative with respect to x of y = x * + ;js + 6 - 10. Calculate the derivative with respect to x of y = 4 x 2 z : -- 1+a: 128. The above method of calculating derivatives is general but often difficult or laborious. To facilitate the calculation of derivatives, several special rules are in common use. These rules enable one to calculate with ease the derivatives of most ordinary functions. 129. The derivative of a constant is zero. This follows from the fact that a constant can have but one value and therefore cannot admit an increment. That is, the increment of a constant is zero and consequently its derivative is zero. This may be symbolized as (1) | C = 0, where C is any constant and x any variable. 130. The derivative of x with respect to a; is 1. For write x = x. Then As = Az, A* As Arc and Lim -r = 1. (la) 198 LIMITS AND DERIVATIVES 131. Derivative of u re , where u is a function of x and n a positive integer. Write y = u n . y + Ay = (w + Aw) n = w n + nu n ~ l Aw + n ^ n ~ ^ w n ~ 2 Aw 2 * L? + (terms containing Aw 3 or higher powers), Ay = nw"- 1 Aw + n(n ~ l \"- z Aw 2 Lf + (terms containing Aw 3 or higher powers), Ay , Aw . n(n 1) A Aw -T-* = mi 71 " 1 -T \- ^ tn '- W n ~ 2 AW -T Ax Az |_2 Ax + (terms containing Aw 2 or higher powers). (2) .-. ife-n,^* */.\ f/.\ since all other terms have zero for a limit as Az > 0, 124. This equation is true for all values of n, fractional, negative, irrational or imaginary. If desired the proofs in ^32, 133, 134, 135 may be omitted. 132. Derivative of u n when n is a positive fraction, say n = p/q where p and q are positive integers. Write y = u p/q . yq = u p (Raising to qih power dy ln u p ~ l du dx dx 133. Derivative of u n when n is negative. Write 1 y + Ay = (w + Aw) m *[nl28n, read factorial n. Ay = DERIVATIVES 199 1 1 (u + Aw)" 1 u m u m (u m + mu m ~ l Aw + m | ~ -^w m ~ 2 Aw 2 + + Aw (u-\- Aw) m w m / ,Aw m(m 1) 9A Aw . . . ,Aw Aj, _ -I*"" AS + ~ J * Au Tx+- ' ' + AM A^ Ax w 134. Derivative of u n when n is irrational. Let m be a variable assuming only rational values as it approaches the irrational number n as a limit. Write n = m -\- e, where e > 0, and consider y = u n = u m+e = u m u e , where e is independent of x. Therefore, since w e > 1 indepen- dently of x we have dy d /,. \ d -T- = -r-[ lim U m U e = -r-U n . ax az\ e __o / dx (5) /. ^ = nu 1 , dx since m > n as c > 0. 135. Derivative of u n when n is imaginary. Let n = mi, where i z = 1. Evidently if i is a number it is constant and so far as calculations are concerned offers no new principle. Hence we may write : y = u mi . dy mi , du (6) -r- = miu" 11 - 1 ^ dx dx Equations 2, 3, 4, 5, 6 show that the equation (2) can be used for all values of n. 136. Derivative of the product of two or more functions. Write y = uv, 200 LIMITS AND DERIVATIVES where u and v are functions of x. Then y + Ay = (u + Aw) (v + At?) = uv + wA0 + Aw + Aw A0. Ay = wA0 + 0Aw + Aw Az>. Ay _ Ay Aw . AM As ~ Ax AOJ As ,_, dy dt> . du (7) -T-= u ^r + v-r' dy dx dx Exercise. Apply formula (7) to show that the derivative of a constant times a function is the constant times the derivative of the function. That is, (la) S*)-c- 137. Derivative of the quotient of one function divided by another. Write ai y = u/v. Where w and v are functions of x: Aw y + AT/ = v + Ac w + Aw w Aw A0 A ^1 -- W-r Ay _ Ax Ax Ax (v + Av) du dv , v-j -- WT- dy <iag dx 138. Derivative obtained from an implicit function. This method will be illustrated by examples. Consider DERIVATIVE OBTAINED FROM AN IMPLICIT FUNCTION 201 Apply the formulas of 129, 130, 131 to each term. The result Solving for -g, (a) ^ - dx ~ a?y ' since -r- = 1. By use of the original' equation y may be elimi- dx nated from this derivative, giving the result in terms of x. Again consider axy + 6 = 0. Applying the above formulas to each term, Hence <w d J.= -y. dx x Find the derivatives of the following: 1. y = 3z 4 -6z+l (formulas (1) and (2)). 2. y = 2x(x 2 + 1) (as a product, u = 2 x, v = (x 2 -f- 1) or write 2 y? + 2 x. 3. y = (x 2) (x 2 + 4z) (as a product w = (x 2), = (x 2 + 4 x) or multiply and use (1) and (2). 4. y = -^- (formula (8)). x -\- 1 5. i/ 2 + 2 z?/ - x 2 = 10 (implicit function 138). 6 - V = 2 ( f nnulas (2) and (8)). 8. ?/ = (1 z 3 )^ w^ where w = 1 x 3 . 9. xY + z 2 + 2/ 2 + 1 = 0. 202 LIMITS AND DERIVATIVES A derivative has, in general, a definite value for an assigned value of the variable. This value is found by substituting the value of the variable in the derivative. 10. y* = 6 x 3 + 4, find ^ for x = 6, and x = 0. ax 11. x 2 + y* = 25, find j- for x at the point (3, 4). dii 12. xy = 12, find ~ for x = 12. Find y from the equation. 13. x/y + 1/x 2 = y/x + y, find dy/dx for a; = 1. 14. y = (a 2 - x 2 )*, find ^ for 3 = a. 15. / = - j- find ~ f or x = 2, and x = 0. (1 a?)* 16. For what values of x is the derivative of x 2 x -f- 5 equal to 0? Equal to 10? (Equate the derivative to and solve for x. Similarly, for all values.) 17. For what values of x is the derivative y = x 3 9 x equal to 0? Equal to 4? dA 18. The area of a circle is A = irr 2 . Find -r- when r = 12. dr 19. From xy = 12, find dy/dx when x = 1, x = 12. 20. From y"- = 4 px, find dy/dx when x = p. 139. Use of the derivative to find the slope of a curve. - The slope of a straight line was defined in 28. The slope of a curve at a point is the slope of the tangent line at that point. Let (1) */=/(*) be the equation of any curve and let PI (xi, 3/1) be any given point on the curve and P (x, y) a variable point in the neighbor- hood of Pi (xi, yi). The slope of the chord PiP is Ayi _ y-yi _ i = - = tan p. Axi x Xi EQUATION OF TANGENT 203 As P moves toward Pi as a limiting position, the chord PiP approaches the position of TPi, the tangent at PI. Then also = -tan*. FIG. 99. It follows that the slope of a curve at a given point and the slope of the tangent to the curve at that point are each equal di/ to the derivative -j- at that point. CM/ It is now easy to obtain the equation of the tangent line to a curve at a given point, if the equation of the curve and the coordinates of the point are known. By 99, Eq. 3, the equation of the line having a given slope and passing through a given point is y 2/1 = m (x Zi). The notation means that x\ is substituted for x in the deriva- tive, and t/i for y also if y occurs in the derivative. 204 LIMITS AND DERIVATIVES For a line tangent to y = }(x) at the point (xi, yi) we must, therefore, have (9) y - yi = \ This is the equation of the tangent line at (xi, y\). 1. Find the equation of the tangent to y z = 8 x at the point where x = 8. dy 4 Here -/- = ax y By use of the equation of the curve, y = 8 when x = 8. Hence dy~] _ 4~| _ 1^ dx] x= s y]y = s 2 Then y - 8 = % (x - 8) or 2y x = 8 is the desired equation of the tangent line. x 2 2. Find the equation of the tangent to y = -^ at the point where x = 2. 3. Find the equation of the tangent to 2 ?/ 2 x z = 4 at the point where x = 4. 4. Find the equation of the tangent to x 2 + y 2 = 25 at the point where x = 3. 5. Find the equation of the tangent to y = x 3 at the point where x = 1. 6. At what angle do x 2 + y z = 25 and xy = 12 intersect ? ./Vote. The angle between two curves at their point of intersection is the angle between their tangents at the point of intersection. See 102. 7. At what angle do x* y 2 = 36 and 2 x 3 y = 1 inter- sect? 8. Show that y 2 = 4 px has two tangents for x = p and that these tangents meet each other at right angles on the z-axis. 9. Show that the tangent line at any point of the parabola y* = 4 px makes equal angles with the line from the focus to MAXIMUM AND MINIMUM VALUES 205 the point of contact and a line through that point parallel to the x-axis. What practical use is made of this property ? 10. Show that the tangent at any point of the ellipse makes equal angles with the lines from the foci to the point of contact. 140. Maximum and minimum values of functions are of frequent occurrence and of much importance. As a simple example, consider the case of a body thrown vertically upward, to find the greatest height it will ascend when its initial velocity is given. The law of the falling body must be combined with the law of uniform motion. The formula is s = v t - \ gt 2 , where s is the height, VQ the initial velocity, g the acceleration of gravity and t the time in seconds from starting. Solving the equation for t gives _ VQ v z 2gs t - g Since v , g, s are essentially positive and t must be real the expression v<? 2 gs must not be negative. Hence s may become so large as to make v 2 2 gs = but cannot increase further without making t imaginary. Hence solving v 2 2 gs = for s gives the maximum value of s to be s = v 2 /2 g. This method is not easy to carry out in most cases that arise in scientific investigations. A much more powerful and general method is given below. 141. Use of the derivative to determine maximum and minimum values of functions. All ordinary functions cf one variable admit of graphic representation. For this reason geometrical problems are convenient and sufficient to illustrate the use of the methods. These can be immediately transferred to any field of science by the medium of graphic representation which is of almost universal application. Definitions. A maximum ordinate of a curve is one which is algebraically greater than ordinates on both sides of itself however near these ordinates be chosen. 206 LIMITS AND DERIVATIVES A minimum ordinate of a curve is one that is algebraically less than ordinates on both sides of itself, however near they are chosen. In Fig. 100, y and y" are maximum ordinates and P and P" are called maximum x points of the curve. Simi- larly, y' and y'" are minimum ordinates. Let y = f(x) be any con- tinuous function of x and FIG. 100. suppose that its derivative is continuous and one-valued in the region considered. Then if 2/1 = f( x i) is & maximum, we shall have (1) /(*i ~ h) </(#i) > /(#i + h), however small h is chosen. Similarly, for a minimum, It follows from the continuity of the derivative and the definition of maximum and minimum, that if x\ corresponds to dii a maximum of the function, the derivative ~r = f'(x) will be ft T positive for values of x in the interval x\ h to Xi, negative in the interval x\ to x\ + h and zero for x = Xi] this is easily seen from Fig. 101. For evidently the slope of the tangent is positive along the arc APi, negative along the arc P\B and zero at PI. Similarly if x\ corresponds to a mini- FIG. 101. di/ mum the derivative ? = f'(x) will be negative for values of x in the interval Xi h to x\. positive in the interval Xi to Xi + h INCREASING AND DECREASING FUNCTIONS 207 and zero for x = x\. In the figure it is seen that the slope of the tangent is negative along the arc A'P 2 , positive along the arc P 2 B' and zero at P*. A necessary condition that y = f (x) shall be a maximum or a minimum value at x\ is that It must be noted that this condition is not sufficient for it may hold at points where the function is neither a maximum nor a minimum as can be seen in the annexed figure. It is now necessary to determine when the condition above gives a max- imum, a minimum or neither. This can be done by the use of the inequal- ities (1) and (2). The steps for deter- ' mining maximum and minimum values of a function are: 1. Calculate the derivative with re- FlG - 102 - spect to the variable, and eliminate the dependent variable if it occurs. 2. Equate the derivative to zero, giving f'(x) = 0, if x is the variable. 3. Solve the equation in (2). Call its roots critical values of the variable. Denote them by Xi, x%, . . . , x n . 4. To determine whether any one of these, say x i} gives a maximum or a minimum, choose a convenient value of h and apply the inequalities (1), (2). 5. Having determined which critical values give maximum and which minimum values, substitute the critical values in the function and determine the actual maximum and minimum values of the function. 142. Consider y = f(x), a function of x. If f(x) increases in value when x increases and decreases when x decreases. f(x) is called an increasing function of x. If f(x) decreases when x increases and increases when x decreases, f(x) is called a de- creasing function of x. 208 LIMITS AND DERIVATIVES Since, if y = /(x) is a decreasing function of x, the increments di] A i/ and Ax are of opposite signs, -p = /'(x) is negative. A given function may be an increasing function in one interval and a decreasing function in another interval. Thus for a maximum : (1) /'(*! - fc) > /'(*0 = > /' (*1 + ti). For a minimum: (2) /'(*! - Ji) </' (*0 = </'(*! + fc). This method is quite similar in manipulation to the method ex- plained in 141. The student will find this method applicable to many problems when the second derivative is not easily obtained. (See 143 for second derivative.) Consider y 2 = 25 x 2 x*, and find the maximum value of this function. Solving for y, f(x) = y = x V25 - x 2 , f'(x) -T- = /^ (use positive radical), /'(*)] Ji H. V2 25-2 v 25 - 3 2 25 - 2 - 4 2 = 3.54, * 2 = + (/i = .54) = -. (ft = .46) V25 - 4 2 Hence/' (x) changes sign from + to at the point x = 3.54 + and /(x) is a maximum for this value of x. Substitute x = 3.54 in/(x) and determine the maximum value. Let the student test the negative value of x = 3.54. Also solve the problem using the negative radical. Draw a graph of the function and discuss this curve. 143. In general the derivative /'(x) of /(x) is itself a func- tion of x and will have a derivative with respect to x. The derivative of /'(x) is the second derivative of /(x) and is denoted by the symbols -|, -7-5, /(x) or/"(x). MAXIMUM AND MINIMUM VALUES 209 In the neighborhood of a maximum point the function f(x') of y = f{x) was shown to be first positive, then zero and nega- tive for increasing values of x. It is, therefore, a decreasing function of x. Therefore if Xi corresponds to a maximum value f 2/ /( x ) we must have /'(%) = and f'M < 0. At a minimum point we must have in a similar way /'(*,) = 0. /"(*i) > 0. These inequalities and equations can be used for determining maximum and minimum values of functions instead of the methods of 141, 142. Some illustrative examples will now be given. 1. Solve the problem at the beginning of 140 by use of the derivative. Write s = VQ[, | gP. Calculate the derivative of s with respect to t, ds # = * - 9t- For a maximum this derivative must be zero. Hence v - gt = 0, t = v /g. This is the critical value of t. Calculate the second derivative This is negative for all values of t, since g is independent of t. The function, therefore, has a maximum for t = v /g. Sub- stituting this value of t in the function gives which is the same result as obtained before. 210 LIMITS AND DERIVATIVES Instead of using the second derivative we may choose a value of h and try to satisfy inequality (1) 141. Thus, let h = e, a small number, and substitute t = e and t = + e in 9 Q the original function and compare the result with that for t = v /g. 20 The last value is greater than the first two for all values of c, however small. Then the last value of s is a maximum. This solution illustrates both methods of procedure. 2. Consider the problem: To prove that of all rectangles having a given area the square has a minimum perimeter. Let A = xy, where x, y are the length and breadth of the rectangle respectively. The perimeter is P = 2x + 2y, dp -9 + 9 d y dx~ = 2 dx' 2 + 2^ = 0. dx dij To eliminate -? take the derivative of A = xy t dx dA dy -=y + x - =0 , since A is constant. dy _ _ y. dx x Substituting in -T- = gives 2 - 2 y/x = or x = y. MAXIMUM AND MINIMUM VALUES 211 That is the length is equal to the breadth and the rectangle is a square. From A = xy and x = y x* = y* = A and x = y = VA, critical value of x. Choose x = VA h, whence by A = xy, A y = p= -- VA - h Hence P = 2x + 2y = 2 \VI -h + -^ - 1 > 4 VZ. L v A hJ Choose x = VA + h, whence by A = xy, A ~ Vl+fc* Hence P = 2x + 2y = 2 |~VZ + h + -7=^ 1 > 4 VI. L V A + hj These results show that if x 9^ y the perimeter is greater than when x = y and the minimum value of P is 4 VZ. To apply the method of 143 we have dx z dx 2 A d?y 2A == x' W = ^' Substituting the critical value, x = + \/A, gives d 2 P 4A1 4A -5-5- = T- = 7= > 0, since A > 0. dx 2 x 3 ] Z =VA VA 3 This test shows that the value x = VA makes P a minimum and consequently from the condition above x = y, the rectangle is a square. Both methods lead to the same conclusion. 3. What are the dimensions of a tomato can of capacity 63 cu. in. of such form as to require a minimum of material, no allowance being made for seams? Write (1) V = itfh = 63, 212 LIMITS AND DERIVATIVES where r is the radius and h the height of the can. Also, (2) A=2Trr 2 + 2irrh, where A is the area of the total surface of the can. Then dA , dh _ -j- = 4 TIT + 2irh + 2irr-r= 0, dr dr for a minimum. This becomes (3) 2r + h + r^ = Q. From (2), dfc 2 A (3a) -3- = --- dr r Substituting in (3) 2r + /i-2fc = or 2 r = h. From (1) by substitution 2 irr 3 = 63. Whence r = 2.15 (critical value of r) and ft = 4.30. From (3), ... d 2 A dh.- d% (4) -j-r = 4ir + 47r-7- + 2irr-7v dr 2 dr dr 2 ,, N d 2 ^ 6/1 From (3a), ^ = - r Substituting values of -T-, -ri, r and /i in (4), ... dM. (5) ^ = 47r - r and a minimum of A is indicated. Substituting values of r and h in (2) gives A =87.1 sq. in. as the minimum value of A. By substituting r = 2, r = 2.3 for r in the expression for A, remembering h = 2 r, gives for A, respectively, the values A =88 and A = 88 , MAXIMUM AND MINIMUM VALUES 213 both of which are greater than the value obtained above. Thus again the two methods agree in their results. Remark. Hereafter the student may solve a problem by one method. He may choose which method to use. In general if the second derivative is easily obtained that method will be best. 1. Show that of all rectangles of given perimeter the square has a maximum area. 2. Show that of all rectangles inscribed in a circle, the square has a maximum area. Note. Assume the radius of the circle equal to r. Consider the geometric properties of the figure and form an expression for the area of the rectangle using a variable dimension and solve the problem. 3. Show that of all triangles of a given base and perimeter the isoceles triangle has the maximum area. 4. Find the dimensions of a cylinder of maximum volume that can be inscribed in a cone of radius 10 and altitude 20. 5. Find the dimensions of a cone of volume 3000 cu. ft. that shall have a minimum curved surface. 6. A fireplace is 2' deep and 4' high. Find the length of the longest straight pole that can be pushed up the 1' chimney. 7. Find the lowest point (minimum ordinate) of the curve y = x 2 + x - 6. x -\- 6 8. Does the curve y = have maximum or minimum ordinate? 9. A carpenter has 108 sq. ft. of lumber. Find the dimen- sions of the box of maximum capacity (lid included) that he can make, making no allowance for joining. 10. A rectangular sheet of iron is 12" x 18". Find the dimensions of the rectangular pan of maximum capacity that can be made by folding up the edges. 11. Find the legs of the right triangle of maximum area that can be constructed on a hypotenuse of 24". 12. Find the dimensions of a right circular cylinder of maxi- mum volume that can be inscribed in a sphere of radius 12. 214 LIMITS AND DERIVATIVES 13. In problem 5, 28, use the three points corresponding to the prices $12, $18, $24, of the profit curve and determine an equation of the form y = ax z + bx + c. From this equation determine the maximum profit and the price corresponding. 144. Use of the derivative to define motion. We are familiar with "speed of a train" or "rate of walking" of a per- son. It is customary to say the speed of a moving body is the distance traveled divided by the time required to travel the distance. This is only an approximate expression of the idea. What is meant is the number of units distance covered, divided by the number of units of time required, is the numerical measure of the average speed over the distance during the interval of time. If speed is not uniform during the interval it will be more than the average speed during part of the interval and less than the average during a part of the interval. If, at a given instant of time, the speed of a body becomes uniform, then the distance As, passed over in the time A, is the instantaneous speed at the given instant. We shall write, therefore, Instantaneous speed = -r- under the above assumptions. Consider the example of a train moving as follows: The train moves 40 mi. during 1 hr. " . " " 25 " " the first | hr. It (( II 10 <( " <( i ft " " " 5 " " " 5min. it tt tt 11 .. u (( i The average speed for each interval is as follows: For 1 hr. 40 mi. per hr. " first Jhr. 50 " " " \ " 52 " " " 5min. 60 " ti tt i n gg it tt Which of the above values of the speed is probably nearest the actual speed at the beginning of the hour? USE OF THE DERIVATIVE TO DEFINE MOTION 215 Suppose the interval of time be decreased toward zero as a limit. The speed will also approach some value as a limit. As The limiting value of the ratio -rr , as Ai 0, is the instan- taneous speed at the beginning of the interval, A. But we know that if s is some function of t, the limit ,. As ds htn -TT = 37 At^o Ac at is the derivative of s with respect to t. That is, the instantane- ous speed is the derivative of s with respect to t, when s is regarded as a function of t. It is noted that so far as the mathematician is concerned s and t are merely two variables related in some way and that he may with equal propriety regard dy/dx as the instantaneous speed of y with regard to x, or re-rate of y. Linear acceleration is defined as the rate (speed) of change of speed. Therefore linear acceleration may be written do . d fds\ cPs acceleration = = = , where s is regarded as a function of tune, I. . 1. A circular plate 6" in diameter is expanding by heat so that its radius is increasing at the rate of 1" per sec. At what rate is the area of the plate increasing? Write (1) A=irr>. rru Then -3r J at at dr Substituting r = 3 and 37 = 1 from the problem, dA -57- = 6 T sq. in. per sec. 2. The edge of a cube is 12" and is increasing at the rate of 2" per sec. At what rate is the volume increasing? The surface? 216 LIMITS AND DERIVATIVES 3. At what rate is the area of a rectangle increasing if its sides a, b are each increasing at the rate of c" per sec. ? 4. Water runs into a conical vessel at the rate of 3 cu. in. per sec. The diameter of the top is 12", the altitude is 18", vertex downward. At what rate is the depth of the water increasing when it is 8" deep in the vessel? Note. Consider a cylinder of height unknown and radius equal to the radius of the vessel at 8" from the vertex and of volume 3 cu. in. 5. A man 6' tall walks along a path which passes under a light 12' above the path. He walks at rate of 3 mi. per hr. from the light. At what rate is his shadow changing length when he is 25' from a point under the light? 6. A body moves so that its distance from a fixed point is given by s = t s + 6 t 2 - 10 x + 18. (a) What is the speed when t = 4? When t = 0? (6) What is the distance when t = 4? When t = 5? (c) What is the acceleration when t = 10 ? When t 1 ? (d) Discuss the graph of each of the functions a, b, c, using t as abscissa. 7. Construct on the same axes the graphs of y = t 3 and of the speed and the acceleration of this function. 8. Construct the graphs of s = 16 t 2 and of its speed and acceleration functions. 9. A particle moves in a path so that the coordinates of its position are x = 1 t + t 2 , y = 1+ t + t 2 . Show that the path is a parabola when referred to x, y coordinates and that the speed of the particle is an increasing function of t. 10. The electrical resistance of platinum wire varies as its temperature 6 C., according to the law R = R (1 + aB + &0 2 )" 1 . jr> Calculate -^ and interpret the meaning of this derivative. uu 11. A body moves on y 2 = 12 x. At what rate is it moving parallel to the z-axis if it moves 50' per sec. in the path when x= 10. Note. Calculate the component of speed parallel to the z-axis. EQUAL ROOTS OF EQUATIONS 217 145. Equal roots of equations. From 85 it is evident any integral function in one variable (or unknown) can be written in the form f(x) = k(x - n) (x - r 2 ) (x - r 3 ) . . . (x - r n ), where A; is a constant and n, r z , . . . r n are the roots of the equation, (1) /(*) = 0. If m of these roots are equal, equation (1) may be put in the form, (2) f(x) = k(x - n) m (x r 2 ) (x - r 8 ) . . . (x - r re _ m ) = 0. Call k(x - r s ) (x - r 3 ) . . . (x r n _ ro ) = <f> (x). Then (2) may be written (3) /(*) = (*-r 1 )0(x)=0. Now (4) f'(x) = m(x - rO" 1 - 1 <j>(x) + (x - ri) m 4>'(x} = 0. The highest common factor of /(x) and /'(#) contains the factor (x ri)" 1 . The equation, (5) (x - n)" 1 ' 1 = 0, contains r\ as a root one less time than does f(x) = 0. It is easy to see that if there were no multiple * roots in f(x~) = 0, there could be no highest common factor containing x. If (3) should be of the form, (6) f(x) = k(x- n)" (x - r 2 ) 1 0(a;) = 0, then corresponding to (5) is (x -n} m ~ l (x -r 2 ) z ~ 1 = 0. The reasoning applies for any number of multiple roots. Hence to determine whether an equation has multiple roots and to determine these roots when they exist, we may proceed as follows: (a) Calculate the derivative of f(x). * When the same root occurs two or more times in an equation it is called a multiple root. 218 LIMITS AND DERIVATIVES (6) Calculate the highest common factor of / (x) and /' (x) and call it F(x). (c) Solve the equation F (x) = 0. (d) Each root of F (x) = will occur in f(x) = one more time than in F (x) = 0. 1. Determine the multiple roots of x 6 2 x 2 + 2 x 2 = 0. 2. Determine the multiple roots of 4 x 3 16 x 2 + 52 x 3 = 0. 3. Determine the multiple roots of x 4 16 = 0. 4. Determine k so that the roots of x'* 6 x + A; = shall have its roots equal. 146. Momentum is defined as the product of mass and velocity. Velocity being a vector, momentum is a vector. If only the numerical value of velocity is considered, the product of speed and mass is the numerical value of momentum. In this sense consider (1) M = mo, where m is mass and v is speed. Taking the derivative dv , since mass multiplied by the magnitude of acceleration is the magnitude of force. Equation (2) shows that force is the time rate of change of momentum. CHAPTER XVI SERIES; TRANSCENDENTAL FUNCTIONS 147. A sequence * is called arithmetic if the differences be- tween successive numbers of the sequence are equal through- out the sequence. In other words when every number of the sequence after the first may be found by adding the same number to the preceding number the sequence is called arith- metic. The number added to a number of an arithmetic se- quence to obtain the succeeding number is called the common difference. Thus, the numbers 2, 5, 8, 11, 14, 17 form an arithmetic sequence whose common difference is 3. A sequence is called geometric if each number after the first in the sequence is obtained from the preceding number by multiplying by the same number. The multiplier is called the common ratio of the sequence. Thus, the numbers 2, 4, 8, 16, 32, 64 form a geometric sequence whose common ratio is 2. The law connecting successive numbers of a sequence may be simple or complicated. When this law can be expressed in the form of an equation it is called a recursion formula. 148. If ai, 02, a 3 , . . . , a n is a sequence of numbers of any kind, then (1) ai + 02 + a 3 + + a n is called a series. If n is infinite the series is called an infinite series. The numbers a\, 0%, a 3 . . . are called the terms of the series. * For definition of sequence see 37c. 219 220 SERIES; TRANSCENDENTAL FUNCTIONS The reason for studying series lies in the fact that a number of the functions with which we have to do in mathematics and its applications are most naturally studied by means of the series which represent them. Many problems are of such a nature as to lead quite naturally to series in their solution. For purposes of this course we shall consider only one class of series, viz., convergent series. A convergent series may be defined as follows : If (2) S n = ai + (h + a 3 + + a n is the sum of the first n terms of the series (3) S = 01 + 02 + a 3 + + a + and if S n approaches a definite limit as n increases indefinitely the series S is said to be convergent. This means that by taking the sum of a sufficient number of terms of a convergent series the difference between this sum and the limit of that sum may be made as small as one pleases. A series with a finite number of terms is always convergent if the terms are finite. In (3) S = lim S n is sometimes called the sum of the n *o series. S will also be used to denote the sum of a finite series. 149. Arithmetic series. Let the series be (1) S = a + (a + d} + (a + 2d) + + (a + (n - 1) d), where d is the common difference, a the first term and n the number of terms. Write the same series in reverse order (2) S = (a + (n - 1) d) + (a + (n - 2) d) + Add (1) to (2), (3) 2S = [a + (a + (n - 1) d)] + [a + (a + (n - 1) d)] (4) Call a + (n - 1) d = I. Then on solving for S, and using (4), (3) becomes fK\ C 0+ f (5) S=n~ ARITHMETIC SERIES 221 Equations (4) and (5) are sufficient to solve all problems relating to arithmetic series. Of the five quantities a, d, n, I, S, three must be known to find two for the two equations (4) and (5) are sufficient to determine two unknowns. 1. How many terms of the series 2 + 7 + 12 + must be taken to obtain a sum of at least 100. He^e a = 2, d = 5, S = 100, to find n, and I if needed. From (4) and (5), <*^ = 100, I = 2 + (n - 1) 5. Eliminating I and solving for n gives 163+ n = r^r = 6.2 and +6 4. Since n must be a positive integer and S must not be less than 100, n must be taken equal to 7. Hence I = 32. 2. Interpolate 25 terms between 16 and 36 so as to form an arithmetic series. Here a = -16, I = 36, n = 25 + 2 = 27, to find d, and S if needed. From (4), 36= -16 + 26d By use of (5), d = 2. S = 270. The series may be written as: -16 -14-12-10-8-6-4-2 + + 2 + 4 + 6 + 8 + 10+12 + 14 + 16+18 + 20 + 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36. 3. Find the sum of 13 terms of the series whose first three terms are, 4, 3^, 2f , . 4. Find the fifteenth term of the series, - 1 - T V + T V + i + - . 5. If the sum of an arithmetic series is 500, the number of terms, 10, the first term, 0, find the common difference and the last term. 222 SERIES; TRANSCENDENTAL FUNCTIONS 6. A hundred apples lie on the ground in a straight line, 4' apart. A basket is 4' from the end of the row in the same line. A boy starts at the basket and gathers the apples into the basket one at a time. How fa*r must he walk? 7. A triangular frame is strung with parallel wires \" apart. The first wire is the base of the triangle, the last \" from the vertex. The base is 24" and the altitude 30". Find the total length of all the wires. Will the angles of the triangle have any effect on the result? 8. The equation of a straight line is y = 2 x -j- 4. Ordinates 1 unit distance apart are erected, beginning at the y-axis, and ending at x = 20. Find the sum of all these ordinates. Find the mean ordinate. 9. Find the sum of all the odd integers from 1 to the nth odd integer in terms of n. 10. Find the sum of the first n even integers in terms of n. 11. Can an infinite arithmetic series be convergent ? Why? 149a. Geometric series. Let the series be (1) S = a + ar + ar 2 + + ar n ~ l . Then (2) rS = ar + or 2 + + ar n ~ l + ar n . (3) Subtracting (r - 1) S = ar n - a A /A\ o ar " a and (4) S = r _ 1 ' Call (5) I --= ar"- 1 , the last term. Then substituting (6) ... S -T=T Equations (5) and (6) are sufficient to solve all problems relating to geometric series. Any three of the five quantities a, r, I, n and S being given the other two can be found, for equations (5), (6) are sufficient to determine two unknowns. 1. How many terms of 3 + 6 + 12 + must be taken to obtain a sum of at least 150? Here a = 3, r = 2, S = 150, to find n, I GEOMETRIC SERIES 223 By (5), (6), 7. = 3.2- i ~~ 1 Eliminating Z and solving for n 2- -51, log 51 Since n must be a positive integer we must take n = 6 in order that S shall not be less than 150. 2. Interpolate 6 terms between 8 and 16 so as to form a geometric series. Here a = -8, n = 6 + 2 = 8, I = 16, to find r, and S if needed. By (5), 16= -8r 7 . Whence r 7 = 2 and 7 log r = log 2 (numerically), log r = ^ = 0.0430, r = 1.104 (numerically). But in this case r is negative and its value is r = 1.104. The series is, therefore, S = -8 + 8.832 - 9.751 + 10.77 - 11.89+13.9 - 14.50+16.01. The result shows that the value of r is only approximate and to get more accurate results a more accurate value of r must be used. It is noted here that when r is negative every other term of the series is negative. S was not needed. 3. Find the sum of | + $ + ^V + to ten terms. Note. Find the tenth term and then calculate the sum. Use equations (5), (6). 4. Find the first term of a series if the sum is 500, number of terms 10 and last term 100. 5. If a = 5. I = 400, n = 10 in a certain series, find r and S. 224 SERIES; TRANSCENDENTAL FUNCTIONS 6. If a + b + c + is a geometric series, show that - + r + - is a geometric series. a b c 7. If 1 + 2 + 4 + 8 + 16 + form a geometric series, continue the series to ten terms and find the sum. 8. Show that a + Va6 + 6 form a geometric series if a and b are both positive or both negative. 9. Show that if log a + log 6 + log c + form an arith- metic series, a + 6 + c + form a geometric series. 10. On a false balance a certain object weighs 9 Ibs. on one pan and 16 Ibs. on the other pan. If x is the true weight show that 9 + x + 12 form a geometric series. 150. Special case of geometric series. Let n oo and \r\ < 1 in _ a ar n o 5 1 r (Note that this equation is really equation (6)). Since | r \ < 1, lim r n = 0.* n >o Hence the above equation will become a (1) 1 -r Therefore when in a geometric series | r \ < 1 and n = oo , the sum $ n has a definite limit as n > oo . This fact will be of much use in later work. * The truth of this is evident when we consider any proper fraction, say f, and raise it to higher and higher powers. Thus ? - = f-\ z - f-\ 3 ! = f-\ ^ = ( 2 \ n 3' 9 W ' 27 \3/ '81 \3/ ' ' ' ' 3" \3/ ' ' * ' It is seen that each successive power of f is less than the preceding and that by taking n sufficiently large (|) n may be made as small as we please. To determine n so that (f) n < ^, write n Gog 2 - log 3) < log A < - I- Hence n (0.1761) > 1 and n > 5.7 Since n is an integer, take n = 6. HARMONIC SERIES 225 1. Find the limit of the sum of I +Q + Q+ ' * ' +3^ as n >oo. 2. Find the value of 4.4747 ... in the form of a common fraction. Note. Write g 4 | 47 , 47 r 100 T 10000 T 47 and sum the geometric series beginning withr^. Add the lUU result to 4. 3. Find the value of 11.911911911 ... in the form of a common fraction. 4. Find the limit of i 1~\ i T\5 i 7i i T\a T ' * * > " i + /i ' (i + ft) 2 (i + /O 3 ATofe. Find the sum of the geometric series whose ratio is 1 1 + /T 5. Find the sum of 12 + 9 + 6f + to an infinite num- ber of terms. 0. If | r | < ^, show that any term of a geometric series is greater than the sum of all the terms that follow. 'Note. Let a be the first term and r be the ratio, r < 1. Form the series and sum all terms after a given term, compare the result with the given term. 7. If the sum of ten terms of a geometric series is 244 times the sum of the first five terms, find the ratio. 151. Harmonic series. If the series (1) S = ai + 02 + a 3 + is a harmonic series, then (2) S' = I + l + i + . . . <Zl 2 #3 is an arithmetic series and conversely. 226 SERIES; TRANSCENDENTAL FUNCTIONS Suppose S = a + x-\-bisa harmonic series in which z is to be determined when a and b are given. By (2), a z is an arithmetic series. Therefore 1 _1 = 1 x a b whence 2ab x = a+b This value of x is called the harmonic mean of a and 6. Hence the series a + x + 6 above becomes o S = a-\ . a -J- b In problems relating to harmonic series it is often better to take the reciprocals of the terms and form an arithmetic series. Solve the problem corresponding to the original and then pass back to the harmonic series. There is no general formula for finding the sum of a harmonic series. 1. Find the harmonic mean of 3 and 7. 2. Form an equation of the third degree whose roots are in harmonic series, the smallest root being 3 and the next largest 4. See 85. 3. In the equation of Ex. 2, substitute l/y for x and deter- mine the roots of the resulting equation. Do they form any kind of series that you know? 4. Show that the geometric mean of two numbers is the geometric mean of their arithmetic mean and harmonic mean. Note. Let a and 6 be the numbers. Form the means indicated and compare as required in the problem. Arithmetic mean of two numbers is half their sum. Geometric mean of two numbers is the square root of their product. 152. Convergence of series. So far as this course is concerned a series must be convergent to be useful in solving CONVERGENCE OF SERIES 227 problems. Only a few standard series which are in common use for studying certain functions that are of great importance in the applications of elementary mathematics will be treated. Only convergent series can have finite and determinate sums. Consider The graph should be drawn to show the curve on both sides of x = 1. A discontinuity occurs at x = 1. The value of y is oo for x = 1. By long division (1) I -^=l + z + z 2 + .T 3 + This is a geometric series having x for its ratio, and having an infinite number of terms. The sum of the series (1) is, by 150, for I x | <1, 1 -x which is precisely the function itself. This shows that (1) is an identity for values of \x\ < 1. When x = 1 both sides of (1) become infinite. When x = 1 the left member is \ while the right member is or 1 according as n is even or odd re- spectively. For x > 1 the left member is finite while the right member becomes infinite. The series on the right in (1) can represent the function on the left only when | x | < 1. For such values of z a finite number of terms will give an approximate value of the function. For values of | x | > 1, a different series can be written which will represent the function. Thus The last series is a geometric series whose ratio is - and therefore x has a definite sum for values of \x\ > 1. A finite number of 228 SERIES; TRANSCENDENTAL FUNCTIONS terms of the series in (2) will give an approximate value of the function. The series (2) does not converge if | x \ = 1 and therefore for such values of x cannot represent the function 1 l-x We shah 1 study functions which can be represented as func- tions of the variable only in the form of series. 153. To use series in the study of functions it is necessary to be able to determine in a given case whether the series is con- vergent. The two following simple tests will meet present needs. (a) Comparison test. Suppose (1) S = Ui + W2 + U 3 ' ' ' + U n 4- Wn+1 + is an infinite series and let (2) S' = V, + V2 + V 3 + + V n + V n+l + . . . be an infinite series known to be convergent. If, beginning at any term of S, the terms of S' and the remaining terms of S can be paired off in such a way that every term of S is less in abso- lute value than its mate in S', or at most equal to it, the series S is convergent. For suppose | U k | < | Vi | , | U k+ i | < | 2 !,.-., | U k+p | < | fljH-i !, - ... for all values of p, where k is finite. It is evident on adding these inequalities, Since S' is convergent it follows that S must be convergent. 154. To employ the comparison test it is necessary to have several standard convergent series for comparison with unknown series. Let (1) S g = a + ar + ar 2 + + ar n ~ l + ar n + , where \r\ < 1. This series was shown in 150 to be convergent. Let (2) s,_i + + + ... ++.... SERIES 229 If p > 1, S p is convergent. For, write ! + !<!- J_ 1__I_|_1 + 1<1- 2? ' SP 2 P 2 P ~ 1 ' 4 P 5 P 6 P 7 p 4? J_ 1-j.l.J., ,+J_< 8 8 p 9 P 15 p 8 P Adding these inequalities, 1 +-L + _L + I A n 1 I O t I This is a geometric series whose ratio is x ? and is convergent if that is, if p > 1. If p = 1, ^zi = 1 and the series does not converge. Series (2) is known as the p-series. Let (3) S = l + l + l + l + . ..+!+.... ( Now l + i>i; * + ! + * + !>! ---- Therefore By taking n sufficiently large S n may become larger than any pre-assigned number M. The series does not converge. This series is called the harmonic series. By comparing a given series with one or another of (1), (2), (3), the convergence or non-convergence of a number of series can be determined. 1. Prove that ifS = Wi + W2 + w 3 +- +u n + does not converge, and if S' = Vi + v 2 + ?'s + + V* + is such that Vi > u\; v z > u z ; v 3 > u 3 ; ... ; v n > u n + the series S' does not converge. Consider positive values only. 230 SERIES; TRANSCENDENTAL FUNCTIONS 2. Determine the convergence of: a. l+ + + (use p-series, p > 1). b. ^ p: + ~ o + o~~J (compare with a geometric series). c. 1 H -7= H -?= + (use p-series, p < 1). 155. Ratio test for convergence. Let (1) S = Ui + M2 + W3 + ' ' + U n + W n+ i + . If Write lim I- -I = k < 1, where k is a fixed number, the series n >oo \ Un I \ (1) is convergent. - = n or - = r 2 or w 3 = r n or Now if a common value can be assigned to the r's so that the above inequalities still hold we may add and obtain the result, + + u\r n + Mir n+1 + If, further, | r \ < 1 the series is convergent. For the right side of the last equation then becomes a geometric series whose ratio is less than one. If | r \ > 1 the terms of S increase with n and the series does not converge. If | r \ = 1, the test does not give reliable results, and some other test must be employed. SERIES WITH COMPLEX TERMS 231 1. Test for convergence 1 . Note. The nth term is 2. Test for convergence 1 1 1-2-3 1 1-2-3 n 1-2-3 ' 1-2-3-4-5 1.2.3.4.5.6.7 ' 3. Test for convergence the series of Ex. 2, 154. 156. Series with complex terms. The real parts of all terms may be separated from the imaginary parts and arranged in a series. The imaginary parts may also be arranged in a series by themselves. Consider S = (fli -f- bii) -{- (#2 ~l~ bzi) -f* - - -f" (fl n 4~ b n i) -J- . Call X = a! + 02 + + a n + and F = &i + 6 2 + + b n + - . Since \S\= Vx* + F 2 , evidently S is finite if both X and F are finite. But X is finite if the series ai + a + + On + is convergent. Similarly, F is finite if the series bi + &2 + + b n + - - is convergent. Hence the con- vergence of the series of complex terms follows. The diagram illustrates the nature of a series of complex terms. Y FIG. 103. 232 SERIES; TRANSCENDENTAL FUNCTIONS n 4. i\z (\ 4. 7 *\3 1. Test l + i+ \ o + T- r^ + ' ' ' for convergence. 1 & L 4 a Note Q 1 , , 1 + 2J-1 , -- ~"~ Y 1 - 1.2-3 y - i I 2 f 2. For what values of x and y is the series ( g + *y) 2 i - convergent. 157. Expansion of functions in series of powers of the variable. A series each of whose terms contains a power of the variable is called a power series. When the powers increase with n the series is called an ascending power series. If the powers decrease with n the series is a descending power series. A series containing only positive integral powers of the variable is called an integral power series. To expand a function it is first assumed that the expansion is possible. If then the coefficients of the terms can be deter- mined, the expansion is known to exist. But the resulting series must be tested for convergence before it can be used in calculations. There is but one power series that can represent a given function in the interval of convergence. By interval of convergence is meant the aggregate of values of the variable for which the series is convergent. If the interval is continuous over a finite region it is sufficiently designated by giving the bounding values. 158. Consider F (x) any continuous function of x having derivatives of all orders which are continuous in a given interval. Let (1) F(x) = A + Bx + Cx z + Dx 3 + Ex* + + Nx n + - EXPANSION OF FUNCTIONS 233 be assumed to hold for a certain interval of x. The values of the coefficients A, B, C, . . . are now to be determined. Calculate: (2) F'(x) (3) F"(z)=2C+2.3Z)aH-3.4#z 2 + / Nn(n-l)x n ~ 2 -] (4) F'"(x}*= 2-3D + 2.3-4k + + Nn\(n - 1) (n - 2) x n ~ 3 + From the form of the above equations it is evident they must hold for x = 0, since A, B, C, . . . are finite constants by hypothesis, provided x = is in the region of convergence of (1). Therefore, F(Q) =A, F'(0) =B, F"(0) = 2C or C = ^5, <u F'"(0)=3.2Z> or D = - Substituting these values of A, B, C, . . .in (1) gives (5) F(*)=F(0)+F / (0)x + ^^a? + ^^^+ _F^O)_ ^ X In this form F (x) is said to be expanded about the origin. If it is assumed that (!') F (x) = A + B (x - a) + C (x - a) 2 + D (x - a) 3 f +#(*- a)" +'-, * By F'"(x) is meant the derivative of F"(x). F"(x) is the derivative of the second order of F(x), F'"(x) is the derivative of the third order, etc. 234 SERIES; TRANSCENDENTAL FUNCTIONS then (2') F' (x) = B + 2 C (x - a) + 3 D (x - a) 2 + (3') F" (x) = 2 C + 2 3 D (x - a) + . (4') F'" (x) = 2 3 D + terms in (x - a). In these equations put x = a, and solve for A, B, C, . . . as above. Substitute the values of A, B, C, ... in (!') and obtain (50 F(x) =F(a) + F'(a) (x - a) + ^^ (x - o) . F'" (a) , + -273-(*-) 3 + " ' In this expansion, a must lie in the region of convergence of (I/). In (50 F (x) is said to be expanded about the point a. If in (50, (x + a) be put for x there results ut \ -pit i i\ (6) In (6), exchange places with x and a and the result is " '" (7) Each expansion must be tested for convergence to be sure it can be used in calculations. 159. Consider the binomial series. Assume (1) (a+x} n =A+Bx+Cx i +Dx*+Ex*+ - +Nx n + - - , where n is any number whatever. Now, by 158, _ , l J. n (n - 1) (n - r + 1) 1.2 ..... r , . Hence on substituting in (1) (2) (a + x) n = a n + na n ~ l x + H ^ ~ -a n ~*x* + n(n-l)(n-2) (n - r + 1) + - 1.2-3 ..... r -a~r THE BINOMIAL SERIES 235 If n is a finite positive integer the series (1) and (2) are finite. If n is not a positive integer the series are infinite series and (2) must be tested for convergence. We shall apply the ratio test. The (r + l)st term is n(n-l) . . . (n-r + _ the rth term is 1.2-3 u - n(n-l)(n-2) . . . (n-r + 2) .. 1-2-3 (r - 1) the ratio n r + 1 The limit of this ratio, as r > oo , is Hence the series is con- a X vergent if : - = k < 1, that is, if | x \ < \ a \ . U 1. Find the value of V31. Write V3T = V36 - 5 = (36 - 5)* = fi/1 _I A \ 6(72-5) \ 2*36" r / 72 Again = (approx.). Note. More accurate results may be obtained by calcu- lating more terms of the series. These expansions are conver- gent. For in each case the second term of the binomial is less than the first, that is, & < 1 and A < 1, which correspond to | x | < | a | in the formula (2) above. 2. Find A/IO = v / 8T"2. 4. Find V3. 3. Find \/18 = v'lG + 2. 5. Find ^ 236 SERIES; TRANSCENDENTAL FUNCTIONS 160. Consider the expression: (I) , W n(n - 1) (n - 2) . . . (n - r + 2) x*- 1 1-2-3 ..... (r - 1) n^ 1 ~* Taking the limit of each term as n > oo the series becomes Equation (2) defines the function / (#) for values of z for which the series converges. It is to be noted that the properties of the function are derivable from the series. By the ratio test = 1-2-3 ..... (r + 1) u r tf _ 1-2-3 The limit of this ratio as ~r > oo is 0, 41, I, if x is finite. Hence series (2) is convergent for all finite values of x and /(x) is a continuous function for all finite values of x. The function f(x) will be called e x (exponential of x) so that (2') f -i + x + *+*2+... + T -f- -+.... When x = 1, there results e=l+l+$+|+...= 2.718281 .... Eleven terms are necessary to obtain this value. The student should make this calculation in full. 161. Theorem on logarithms. Log 6 y = log& a log a y, where a, 6. y, satisfy the definitions of 19, 20. Let logo?/ = u and lo&y = v. Then y = a u and y b v . Therefore a w = 6". a = 6 V/U . log & a = v/u = log& 2//log y. That is, log& y = log a y log& a. THE SERIES 237 This is exactly the theorem. This equation makes it possible to change the base of a system of logarithms. The two bases in common use are 10 and e = 2.718281 .... Example. Let b = e, a = 10, y = 25 and logio 25 = 1.3979. Tofindlog e 25. By the above theorem write, log 25 = log e 25 - logio e, whence log e 25 = logic 25 ,/logio e or loge 25 = 1.3979/0.4343 = 3.193. 1. Using the tables for finding logarithms to the base, 10, by above method calculate the logarithms of 20, 15, 35, 75, to the base e. The number, log e 10 = 0.4343, which as a multiplier would change log e y to logio y, is called the modulus of the system of logarithms to the base, 10. 162. The series (2'), 160, may be used to obtain the deriva- tives of exponential function and the logarithmic function. Write (1) y = e u , where u is a function of some variable, say x. Then y -f- Ay = e u+Au = e u e Au , Ay = e u (e AM - 1), / (Ax) 2 (Ax) n \ i A i j_i* i 2 'l.9. . *j -* J L\y ..N / Ax /Aw . Ax Aw (Aw) n-1 Aw . V ~~ 1 AT * A-r ' 'l.9. .A' j- = dx dx' since all terms after the first vanish with Ax. Now by (1), (3) log e y = w. By rearranging (2) and using (1), (A\ du _ ^ dx _l dy ~ e u dx~ y 238 SERIES; TRANSCENDENTAL FUNCTIONS Therefore (5) I 1 *'-*- If a is the base of the system of logarithms, multiply both sides of (4) by logo e, 161, and there results d, 1 ^Iog a y = log 6.-. 1. From a table of logarithms to the base e, construct the graph of the equation y = log e x. 2. From a table of logarithms to the base 10, construct the graph of the equation y = logio x. Compare the graphs of this exercise with the graph of Ex. 1. Discuss both. 3. Construct the graphs of y = (? and y = log x on the same axes to the same scale. Note the positions of the curves. Each is the image of the other reflected in a mirror placed at an angle of 45 with the z-axis. Such functions are inverse to each other. See 70, 71, graphs of sin x and arc sin x. 4. Construct graphs of y = x z and y = x% on same axes and compare results with above. 163. It is often convenient to use the logarithmic function in calculating derivatives of algebraic functions. Consider- spin * = V (2) log* y = m loge x - n log e (1 - x). 1 . 1 (4) y dx x 1 x dy _ y [m (m ri) x] dx~ x (1 x) (m n) x m mx m ~ l 1. Find the derivative of y = x/(l x}, by above method. 2. Find the derivative of y = (1 a 1 ) 2 , by above method. 3. Find the derivative of y = e?x n , by above method. LOGARITHMIC SERIES 239 4. Find the derivative of y = u v , where u and v are functions of x. Save result as a formula. 1 + x z 5. Find the derivative of y = r -- % JL ' 6. Calculate the tabular difference for logarithms of numbers between 120 and 121. (Use Eq. 5, 162.) 164. It is now quite easy to derive a series by means of which the logarithms of numbers may be calculated. Using Eq. (7), 157, write (1) /(* + o) = log, (x + a) = f (x) + /' (z) a + f -- a 2 2-3 The derivatives are f(x) = \Og e X, Substituting in (1) gives: a a 2 a 3 /O\ 1 / I \ 1 I I If x = I this becomes, (4) Write also log e (1 o) = a -^ ~- -j 3S o 4 /< i _\ / (5) .-. loj Now put a = ~ - in Eq. (5) and transpose L tn ~p 1 (6) Iog.( 240 SERIES; TRANSCENDENTAL FUNCTIONS This series converges .rapidly and three or four terms are suffi- cient for calculating logarithms of ordinary numbers. 1. Test series (6) for convergence. 2. In (6) put m = 1, 2, 3, in succession, using three terms, multiply each result by logio e = 0.4343 and compare the results with the logarithms of 2, 3, in the tables. 3. Given logio 10 = 1, calculate by use of above series log 11. 4. Given logio 3, logio 4, taken from the tables, calculate logio 13. e - = l + ^ + M 2 + M 3 *+ .. . = !-*-+ ~ 4 2 ' 2-3.4 (/y3 X ~2T? + 3 ' 2.3.4.5 where i? = 1. Now write yvc _ p-ix (~\} f(r W J V*J ^ o and (2) ft&^-^f-' Squaring these by actual multiplication and adding gives [/(z)] 2 + L/i ( x )] 2 = I- Compare this result with equation (1), 48. Again carry out the work and obtain from (1), (2) above, _ f>-i(x+y) Compare this result with equation 24, 53. These two relations are sufficient to show that/ (x) and/i (x) are the sine and cosine functions. But which is sine and which is cosine is not yet determined. Substituting x for x in/ (x), the result is /(-*) = -/(*)- * Here as in a previous theorem we have assumed i to be a symbol treated as a number. DERIVATIVES 241 Therefore / (x) is an odd function like the sine function. By the same substitution, /i(-*0 =/i() and fi (x) is an even function like the cosine function. It is safe to conclude that / (x) = sin x, /i (x) = cos x. The right members of (1) and (2) are known as Euler's expres- sions for the sine and cosine functions, respectively. 166. The derivatives of the sine and cosine functions can be obtained by use of (1), (2), 165. Write e iu _ e -iu mi. dy d . 1 1 nen ~r~ = T~ sin u = =-. dx dx 2i\ dx _ (e iu + e~ iu ) du ~2 dx du = COSM-j-- dx (1) .-. Similarly sinu- dx y = dy du COSU -j-' dx giu _j_ giv ' -iu^ U dx d dx dx dx COBU 2 _ (e iu e~ iu ) du 2i dx /rkN d . du (2) .*. -j- cos u = sin w 3 dx dx Tt r x sin u , . By use of tan u = derive cosw / \ d , du (3) -r- tan u = sec 2 u 3 dx dx 242 SERIES; TRANSCENDENTAL FUNCTIONS . d , du 1. Derive -r- cot u = esc 2 u -=- ax ax n TT . 1 , . a* du 2. Using sec w = - . derive -r- sec u = sec w tan u -r- cos u ax dx 3. Using cscw = - , derive -3- cscw= cscwcotw^-- smu dx dx 4. From y = sin 2 u, find dy/dx. 5. From y = sin 2 u = 2 sin w cos u, find dy/dx and show that cos 2 M = cos 2 u sin 2 w. 6. From y = 2 sin 2 w + 3 cos u, find dy/dx. 7. From ?/ = tan u-\-u, find dy/dx. 8. From y = log sin w, find dy/dx. 9. From T/ = log sec u, find dy/dx. 10. From y = (a sec 2 u + 6 cos 2 w) 3 , find dy/dx. 11. From ?/ = e sinx , find 12. From y = tan 2 w + ^ find dy/dx. vi 13. From ?/ = sin u cos v + cos u sin 0, find dy/dx. Regard each term as a product of two functions. 14. From y = - ^~ , find dy/dx. cos2w 15. From y = sin 3 u + 6 cos 2 u, find dy/dx. / 3> 16. From y = 2 sin ~ 3 tan ~, find dy/dx. 2i & 167. To expand sin w and cos u in power series, write by 157 (5) d 2 . 1 , -, -r-^sinw /, XJ./N . .-..a. . au Ju = o oi (1) /Cw)=sinw=sinO+-r-sinw u-\ -- ^ - w 2 + . au Ju = o ^ Now -r- sin u = cos w. aw d j aw 2 aw sm u = cos u = sin u, a 74 d , , 3, sin u = 3- ( cos u) = sm w. du* du PROBLEMS 243 Setting u = in each of these and substituting the values in (1) gives u 3 u 5 sn u = u 2.3^2.3.4.5 1. Test the last series for convergence. Compare the last series with the real part of the expansion of e**, 161. 2. In a manner similar to the above expand cos u in a power series. Compare the result with the imaginary part of the expansion for e ix , 161. Test the last series for convergence. Note in using derivatives and series of the trigonometric functions the variable must be expressed in radians. TT 3. Using the series obtained for sin u, put u = ^ (?r = 3.1416) and calculate the sin ^ = sin 30. Compare the result with the value in a table of natural sines. 4. Using the results obtained above can you now prove that e = cosx -f- ismxt 5. Plot graph of and discuss y = e x \ find slope at x = 1, x = l. 6. Plot graph of and discuss y = e~ x \ find slope at x = 1, x = \. x = rO r sin 0, 7. Plot graph of and discuss , ( y r r cos 6, r constant, x, y, the coordinates of points on the curve for values of 6. This curve is the cycloid. Read up in an encyclopedia on this very remarkable curve. 8. Find the slope of the cycloid at = 0, 6 = 7r/4, 6 = IT. dy dy I dx Note. Use /- = - / -^- dx d6 / dd 9. Construct and discuss y = sin x + cos x. 10. Construct and discuss y = e~* sin x; find slope at x = ir/2. 11. Find the angle at which y = sin x and y = cos x, in- tersect. 244 SERIES; TRANSCENDENTAL FUNCTIONS 12. Find the equation of the tangent to y = sin x, at x = T/6, X - 7T/2. 13. Construct and discuss r = ae" 6 (polar coordinates). 14. Construct and discuss r = cos 3 6 (polar coordinates). 15. Construct and discuss y = cos 3 x (rectangular coordi- nates). 167a. In certain classes of problems where the formulas to be determined are of the form y = Ax n , it is often easier to de- termine the constants from a graph constructed with the log- arithms of the number pairs instead of the numbers themselves. This can be done by use of a table of logarithms and ordinary coordinate paper. Of such frequent use is this method in certain branches of engineering that special coordinate paper is used. This paper is divided and ruled in spaces propor- tional to the logarithms of numbers as is the slide rule. It is then only necessary to plot with this paper as with ordinary paper. Such paper is called logarithmic coordinate paper. Suppose we have the number pairs, 3 10 36 400 0.477 1 1.556 2.602 Locating the x, y number pairs directly on the logarithmic diagram, we obtain the line AB. To determine n measure DC DC to any scale and 4 C to the same scale. The ratio j-^ = 2 is the slope of AB and is the value of n. The value of A is the number corresponding to 1~4 or 4. Now substituting these values in the proposed equation y = Ax n there results y = 4x 2 as the desired equation. To see that these values are thus determined write the pro- posed equation in the form logy = log A + nlogx. x = 1 1. 5 2 y = 4 9 16 then logs = 0. 176 0.301 and logy = 0.602 0. 954 1.204 LOGARITHMIC GRAPH 245 As log y and log x are the coordinates of points in this method log A is the ^/-intercept of the line and n is the slope. I I 5 A 3 1 2 I So 5 I t \ I / / / / ID / / / / / / _J_ / / A r* O i g 345678 910 2 3456789 100 Logarithmic Scale FIG. 104. The student should procure several sheets of logarithmic coordinate paper and solve the above problem completely. It should be noted that the divisions on this paper are exactly those on the A scale of the Mannheim slide rule. 1. Draw the graph of the above problem on ordinary co- ordinate paper and on logarithmic paper. Interpolate several points on each graph and determine their coordinates from each graph. Which of the graphs affords the easiest and most accurate interpolation? It will be seen that the logarithmic graph has decided advantages. 2. Make the graph on logarithmic paper and determine an equation of the form y = Ax n from the data below and also construct the ordinary graph. x=l 2 3 4 5 = 2 8 18 32 50 246 SERIES; TRANSCENDENTAL FUNCTIONS 3. Construct the logarithmic graph and determine the equation as directed in (2) from the data below. x = 10 30 40 60 80 100 y = SQ 61.5 45 34 26 21.5 Remark. It should be noted that the logarithmic coordinate paper provides directly only for numbers from 1 to 100. The scale is often 5 in. to 1 unit. Each of the two main divisions of the paper provides for numbers whose logarithms have the same characteristic. If numbers from 100 to 1000 are to be represented, the axes, main division lines of the paper, must be renamed. Ordinarily the left margin is named 1, the middle line 10 and the right margin 100. The axis 1 may be called 10 or 100, and the others correspondingly. This is equivalent to changing the characteristic without disturbing the mantissa of the logarithms. To illustrate this the lower margin of the paper with axes renamed is shown in the figure below. Logarithmic Taper 1 or 10 or 100 10 100 1000 101 100 1001 FIG. 105. 167b. Illustrative problem 1. The"electrical resistance of a certain river water when different amounts of solids were in solution was found to be as follows: Resistance (ohms) y = 1000 800 600 400 300 200. Solids (parts in 10,000,000) x = 215 260 340 480 615 860. On the logarithmic paper let K be the point (1000, 100) at first. Lay off the points from the data of the problem; they determine the line BAA 1 . Drop the point A 1 to A 11 and then move A 11 to A m . Draw A m A rr parallel to AA 1 . The new names of the axes appear underscored on the diagram. The point A is the ^/-intercept of the line. For A IU A IV is in reality a continuation of BAA 1 , and A iy read to the new-named axes is ILLUSTRATIVE PROBLEM 2 247 the point where the line crosses the axis originally named 1. Since the graph is a straight line on the logarithmic paper the equation desired will be of the form y Ax n . Proceeding as in the illustrative problem of 34, the values of A and n are A = 300,000 and n = -1.05. The equation is. therefore, y = 300, VBOOOOO \.r 100 Sft - 1000 10 Note. The sliding of the graph downward and to the left is done to avoid having to extend the drawing off the paper to the left and upward. The slope of BA 1 is negative because the line slopes to the left and upward. This will be further explained later. 167c. Illustrative problem 2. Let us determine the equa- tion of problem 4, 36, by a method similar to that of the last section. Talcing logarithms of both sides of the equation gives log p = log A + (K log e) h. 248 SERIES; TRANSCENDENTAL FUNCTIONS This equation is of the first degree in log p and h. This suggests making the graph with a logarithmic scale for p and an ordinary scale for h. It is evident the slope of the line is K log e and the intercept on the p-axis is log A. The point A is 30. Measure AD to the scale 2" to 1 unit and DC to the scale 2000 to 1". = 0.0000625. The Then the slope of CA is slope is also K log e and log e = 0.4343. 0.0000625 Hence Hence the equation is 0.4343 = -0.00144. = 30e-- 00144A . 100 lio 6 o -8000- 2000 4000 Uniform Scale FIG. 107. 6000 The paper ruled to logarithmic scale one way and to ordinary scale the other way, as used in this problem, is called semi- logarithmic paper. ILLUSTRATIVE PROBLEM 249 1. Construct on semilogarithmic paper the graph and deter- mine an equation of the form from the data below. A thermometer initially at 19.3 C. is exposed to the air and the readings taken afterward as follows: Time (seconds after starting) t = 20 40 60] (uniform scale) Temperature C = 19.3 14.2 10.4 7 . 6](log. scale) 2. Construct the logarithmic graph and determine the equa- tion from the following: The amount of water that will flow through a certain length of pipe of different diameters under a pressure of 50 Ibs. per square inch is given below: Diameter in ft. d = l 1.5 2 3 46 Quantity water cu. ft. per sec. q = 4.88j 13.43 27.5 75.13 152 409 3. Given the number pairs: x = 1 2 3 4 2/ = ^ A- 2.7 6.4 Draw graph on logarithmic paper and determine the equation. 4. The following data were observed on a given amount of gas when no heat was allowed to enter or escape: Pressure p = 20.5 25.8 54.2 Volume v = 6.3 5.3 3.2 Plot on logarithmic paper and determine the relation between p, v in the form of an equation. CHAPTER XVII INTEGRATION 168. The derivatives of several types of functions have been considered. Some of the applications of derivatives have been illustrated. It will be profitable now to consider the inverse problem, viz.: Given the derivative of a function, to determine the function. This problem and the process involved are included under the name integration. The sign of integration a /* is I . When this sign stands before an expression it indi- cates that a new function is to be determined whose derivative is the expression under (immediately following) the sign of integration. Integration enables us to solve a great number of new problems in science and geometry. Consider, I 3 z 2 = y? + C, where C is any constant. To prove this equation, take the derivative of the right side and compare with the expression under the sign of integration. This derivative is seen to be exactly 3.r 2 , which proves the truth of the equation. C is called the constant of integration and cannot be determined without further information. The expression x 3 + C is called the integral of the expression under the sign of integration, 3x 2 . In what follows, it is assumed that the integrand* is continuous for values of the variable considered. For convenience of reference the derivatives of a few funda- mental functions and the corresponding integrals are given: * The integrand is the expression under the sign of integration. 250 DERIVATIVES AND INTEGRALS 251 4141 14 414 414 14 8 14 W -I4 -I4 252 INTEGRATION The chief difficulty in integration is to recognize the type dit form of the integrand and the factor to be used as -r To fac- dx tor and arrange an integrand to fit an 'integration formula, it is often necessary to introduce or take out a constant factor. The selection of the formula of integration is a matter of judg- ment and experience. To illustrate, take fxe = I f e 3 *' 6 z = e 3 * 2 + C (formula 7). To verify the result find the derivative of the integral. This is precisely the original integrand xe* x *. A variable factor must not be introduced under the integral sign nor taken from under the integral sign. Consider / (a + 6z 2 ) 3 x. This may be written: 8. . fz* = ? 9. fe 2 ' = 3. f\=1 10. J^ + e*' 2 ) =? 4. 13. 7. + 1) (1 + x) = ? 14. 3a* = ? (multiply) * Each integrand may be multiplied by dx before integrating, if desired. See 176. DIRECTIONS FOR SOLVING PROBLEMS 15. f^|^ = fees- 2 3 z sin 3 z=? J cos 2 3 x J 16. fsin 3 zcosz = ? 23. Cf^. = ? J J tana; 17. f (I + 2x) (x + x 2 ) = ? 24. f csc 3 a; cot x = ? 18. f (1 - cos 2 a;) =? 25. fx cos x 2 = ? 19. /tan z sec 2 z = ? 26. f cos z e sin * = ? 20. f cot 2 a; esc 2 z = ? 21. /?-? . r(l 253 . J (I - xrfx = 22 35. . f(l + 3) (1 - a;) = ? 36. fcos (3 x + 2) = ? -z 2 -z-l) =? 37. f6sin(4z 2 -l)z=? . / (1 - sin x) cos x = ? 38. j(l + tan 2 x) 2 sec 2 z = ? 31. 32. 33. 34. 169. After some practice in the process of integration some easy applications can be taken up. The solution of a problem will consist of the following steps: 1. From the given data, formulate the correct integrand. 2. Select the formula for integrating. 3. Rearrange the integrand, introducing constant factors, if necessary, to make it fit the formula. 4. Integrate the expression. 254 INTEGRATION 170. The function obtained by integrating will have a dis- tinct value for each value of the variable substituted in it. The difference between two values of the integral corresponding to two values of the variable is called a definite integral. Thus if be evaluated for x = 1 and x = 5 and the former value sub- tracted from the latter there will result (2) (5) 4 + C-R 4 +c]=624f. The values x = 1 and x = 5 are the limits of integration, the value x = 5 being the upper limit and the value x = 1 the lower limit. As a brief way to indicate what was done in (1) and (2) above we write /5 T 5 ~[x = 5 *-|+C -624*. Ji = l Now consider the second illustrative problem of 170. If x = and x = 4 be taken as the limits of integration, there will result In the process of subtraction in each of the above cases the constant of integration C was eliminated. This will always occur. When the limits of integration are given it is unneces- sary to determine C for it can be eliminated. An integral in which the undetermined constant of integration appears is called an indefinite integral. All the integrals of 168 are in- definite integrals. 1. Given the instantaneous speed of a point moving in a straight line as a function of the time, t, from some position of reference, v = 1 + 6 P. Find the displacement function and the space passed over during the interval from the beginning of the 5th to the end of the 12th second. DEFINITE INTEGRAL 255 fj? We have v = = 1 + 6* 2 , 144. Hence s = f (1 + 6 Z 2 ) = * + 2 Z 3 + C. This Is the displacement function with C undetermined. The space passed over is the value of the definite integral, 12 = 3336. 2. In the above problem, instead of the given data, suppose that t = 0, when s = Q, and let us solve the problem with these conditions. We have from above, s = t + 2 f 3 + C. For t = 0, s = 0, = + + C and C = 0. Hence s = t + 2 1 3 is the complete space function. The second part of the problem is solved now as s = t -\- 2 * I = 3336 as above. 3. The acceleration of a particle is a = ktf. Find the speed and displacement functions and determine the speed and dis- placement from rest to t = t\. See 144. 4. The speed of emptying a vessel by a hole at the bottom varies with the square root of the depth of the hole below the surface of the liquid. How long will it take to empty a cylinder 3' in diameter, 10' high, if the_opening is 1" in diameter? Given the speed of flow v = V2 gh, where h' is the depth and g = 32.16'. 5. The speed of a body from rest is v = 3 t z 6 t + 10. Find the acceleration at t = and at t = 5. Find the distance passed over from rest until t = 5, t = 10. 256 INTEGRATION 171. Area under a curve. The area under a curve is that portion of the coordinate plane bounded by the curve, the x-axis and the ordinates at the ends of that part of the curve considered. Under this definition portions of the area that may lie below the x-axis are regarded as negative. In the figure PQXzXi is the area under the curve PABQ. An element or increment of area is defined as a narrow strip bounded by the arc 0, the ordinates at its ends and an element Ax of the x-axis. FIG. 108. Now take A A = shaded area + ACB = y Ax + 9 AT/AZ, where CB = Ay and 0^1. If </=/(*) is the equation of the curve, we have AA = /(x) Ax + A/(x) Ax, dA since A/(x) = Ay as Ax 0. VOLUMES OF SOLIDS OF REVOLUTION 257 It follows that /**, C 1 * = y = /(x) \J Xi *S X t is the area under the curve. 1. Find the area under y 2 = 12 x, between the ordinates for x ='0and x = 10. Write /MO /no . _ , v2 r ~\ x = 10 A= \ y= Vl2-x* = ^f =72.5. /0 /0 i>// Jz=0 2. Find the area under i/ = x 2 + 2x 1, between a; = 0, re = 6. x 11 3. Find the area under o + f = 1 and the axes. 4. Find the area under xy = 12 from x = 1 and a: = 12. 5. Find the area between xy 12 and x -f y = 12. Note. The area between the curves is the difference of the areas under the curves between the abscissas of the points of intersection. 6. Find the area under y = sin x, from x = to x = TT. From x = IT to x = 2 ?r. 7. Find the area of the triangle whose vertex is at 0, alti- tude, 16, lying on the x-axis and base 24, perpendicular to the x-axis. 4 8. Find area under y = - from x = 1 to x = 10. * 9. Find area under y = e? from x oo to x = 0. 10. Find by integration the area of a rectangle of base, 6, and altitude, h. 172. Volumes of solids of revolution. Let y = f(x) be a curve. Let any arc of the curve, between two ordinates, revolve about the a;-axis. The solid described is a solid of revolution. If an element of the curve, AB = As, revolve about the axis a thin lamina or slice is generated whose volume is A V = Try* As + 2 irB As Ay = TT [/(x)] 2 Ax + 2*6 Ax A/(x), where = 1, and as Ax >0. 258 Hence and Hence FIG. 109. 1. Find the volume generated by revolving an arc of 'y 6 x + 1, between z = 0, x = 10, about the x-axis. Write _, (6 * + !). /MO = / T< t/0 18 = 12610 TT. 2. Find the volume generated by revolving y = e* about the x-axis between the limits x = 0, x = 10. Also between x = oo, x = 0. THE AVERAGE VALUE 259 3. Find the volume generated by revolving y = - about the C ce-axis from x = 1 to x = 10. 4. Find the volume generated by revolving y = sin x about the re-axis from x = to x = TT. Note. Remember 2 sin 2 x = 1 cos 2 re. 5. Find the volume generated by revolving y = e x/2 e~ x / 2 about the re-axis from x = to x = 4. 6. Find the volume bounded by the two surfaces generated by revolving xy = 12 and x + y = 12 about the re-axis. 7. Find by integration the volume of a cylinder of radius, a, and altitude, h. Note. Consider the cylinder generated by revolving a rectangle. 8. Find the volume generated by revolving y = 8 x about the re-axis, from x = to re = 10. From this result derive the rule for finding the volume of a cone when the radius and altitude are given. 9. Find the volume generated by revolving y = x 3 about the re-axis, from re = to re = 4. 10. Find the volume generated by revolving re 2 + y 2 = r 2 about the re-axis, from x = to re = r. Note. Solve the equation for y, y = Vr 2 re 2 . From the result derive the rule for finding the volume of a sphere of radius, r. 11. Find the volume generated by revolving 9 re 2 -f- 16 y z = 144 about the re-axis. 12. A complete meridian of the earth is an ellipse whose long diameter is 7926 mi., and short diameter 7899 mi. Write, in standard form, the equation of this ellipse and by the method of Example 11, find the volume of the earth in cu. mi. 173. The average value of a function over an interval of the variable. Let y = f (re) be the function and Xi < x < rc 2 the interval. From the figure it is evident the average value of the function is the average of the ordinates of the curve over the interval, and is the altitude of a rectangle whose base is 260 INTEGRATION and whose area is the area under the curve. Therefore, write A /w / A j )fe \ - = y a = - or \ - = 2/o= - I- Xz xi Xz xi \Xz-xi X2 Xi/ This expression gives the average value sought. This is an extension of the idea of average to an infinite number of values. Y FIG. 110. 1. Find the average ordinate of y = sin x between x = and x = TT. Write smx = - = 0.63-K 7T z = 7T Note. This example has important bearing on the theory of dynamos and motors. 2. Find the average ordinate of y = cos x between x = and x = IT. Note. Remember areas below the a>axis are negative. 3. Find the average ordinate of y = cos x between x = = and|. WORK DONE BY A VARIABLE FORCE 261 4. Find the altitude of a rectangle of base 12 that equals the area under y = x z + 6 x + 27. Draw figure. 5. Find the average ordinate of y = a 2 x 2 between x = a, x = a. 6. Find the average ordinate of x + y = 12 between x = and x = 12. 7. Find the average ordinate of y = ^ ; between x = 1 + x andx = |. Between x = 2andz = 1. Between x = 1 and x = 0. Between x = 2 and x = 0. How do you ex- plain these results ? Draw figure. 8. Find the average ordinate of y = e~ x between x = 3 and x = 3. 9. Find the average ordinate of y = cos x + sin x between x = ir/2 to x = TT. 10. Find the average values of y = x, y = x 2 , y = x 3 , re- spectively, from x = to x 10. 174. Work* done by a variable force. Suppose the force is expressed as a function of some variable, t, and that the displacement is expressed as a function of the same variable. Thus let the force be /-*(*) and the displacement in the direction of the force s = F(f). Now the element of work is Aw = As-/ + 8 A/. As, where / is ordinate of P, CP = A/ and 6 = 1 (see Fig. llOa), A/ being positive or negative according as the force is an increas- ing or decreasing function with the displacement. Then * Work is defined as the product of force by the component of displace- ment in the direction of the force. 262 That is, INTEGRATION Aw _ As . ,,As dw ,ds $-*.*. FIG. HOo. Therefore 1. What is the work done by a force f(x) = 10 x 2 in a dis- placement <j)(x) = x, from x = to x = 10? Write 3 ~\ x = 10 - = 3333 units of wor-k. /MO /no = / 10 a; 2 - 1 = / 10 x* = t/o Jo 2. It requires a force of 5 Ibs. to stretch a spring 1". If the force required for any stretch is proportional to the stretch, find the work done in stretching the spring 1^ ft. Note. Write f(x) = kx and <t>(x) = x. DIFFERENTIAL 263 3. The force of gravity varies as the square of the distance from the center of the earth. Find the work required to lift a 1-lb. mass from the earth's surface to a point 500 mi. high. It is given that force is 1 Ib. at the surface and the radius of the earth 4000 mi. 4. Find the work done by a force f = <t>(t) =3 3 8 + l in a displacement s = F(t) = 8 1 6, from t = to t = 40. 5. A force varies inversely as the displacement; when the displacement is 1 the force is 100. Find the work done in a displacement of 100 from a displacement of 5. 6. What work is done in winding, on a windlass, a chain 100' long, weighing 2| Ibs. to the linear foot ? 7. How much work is done by a force that can just roll a barrel weighing 300 Ib. up a smooth incline of inclination 30 with the horizontal? 175. So far we have regarded the integral solely as a function whose derivative is given, that is, as the inverse of the derivative. This viewpoint is fundamental and of great value. In order to realize more fully the power and utility of integration as a mathematical instrument we must take another, though not contradictory, viewpoint and look upon an integral as the sum of infinitesimal elements under certain conditions. This idea is implied in the second form of integrals given in the preceding list of integrals. Let us now consider the problem of 171. We have (1) AA = y Ax + 6 A?/ Ax. The first term on the right, y Ax, is defined as the differential of A or " differential A " and written dA. We shall write dx (differential x) for Ax and instead of the above equation we shall consider (2) dA = y dx. Note that differential A, (dA), is precisely the derivative of A with respect to *, ( -r- ) , multiplied by dx. This prin- ciple is general. It is now seen that y dx is a rectangle inscribed 264 INTEGRATION under the arc AB and that the area under PQ may be con- sidered the limit of the sum of a set of such inscribed rectangles as their width, dx, approaches zero. This idea is similar to the one used in elementary geometry to show that a pyramid is the limit of the sum of a set of inscribed or circumscribed prisms. See Wentworth, P. and S. Geom. (rev. ed.), p. 312, or some other text. We may now write = f^ydx = t/Xi (3) A = I dA = I ydx = I f(x) dx = F(xy) F(XI), tS Xi J X\ *J Xi where f(x) is the derivative of F(x). This example shows that integrals considered from the new viewpoint need no rules or processes different from those developed for the inte- gral as the inverse of the derivative. The only difference lies in the construction and interpretation of the integrand. Example. Let the student revise, according to the new viewpoint, each of the illustrative problems of 172, 173, 174. 176. We shall now apply the idea of summation to the solution of a new type of problem. In the solution we shall need the idea of moment 81, Ex. 5. Note. The centroid of a system of parallel forces is the point of application of their resultant or equilibrant. The equilibrant is equal to and opposite in direction to the re- sultant. The center of gravity of a material body is the centroid of the forces acting between its particles and the earth. With these definitions we now proceed to find the center of gravity of a thin straight rod AB of unit mass per unit length and length 1. Let ab be any small portion of AB of length dx. Its mass and the measure of its attractive force may also be taken as dx, since the density is unity by hypothesis. Taking A as the origin, AB as the z-axis, I as the length of AB and x as the distance from A to some point of ab, we have for the moment of ab about A, CENTROID 265 a b B x dx (1) dM = x dx. If we take the sum of the moments of all such elements as dx > 0, we have (2) M= f l dM = f l Jo Jo The sum of all the forces has the same measure as the mass of the rod. This is, of course, I, which may, for theoretical con- siderations in later work, be looked upon as C l C l T (3) m = I dm = I dx = x\ = I. Jo Jo J The distance x of the point of application of the resultant of the forces from A is equal to the moment M divided by the mass (sum of forces) m or P dx That is, the center of gravity of a thin straight rod is at its center of length, which is what might have been expected from considerations of symmetry. For the value of the above processes in later work the student should thoroughly master them. Let us now find the centroid of the area under the curve, y = f(x) in Fig. 108. The element of mass (force) is now (5) dA = y dx = f(x) dx. The distance of this element from OY (conveniently chosen as the axis of moments) is x. We may now write (6) dM = xdA = xydx = f(x) x dx as the moment of any element [rectangle under AB]. Hence (7) M = I xdA = I xydx = I f(x)xdx = <f>(x 2 )(l>(xi') > Jxi Jxi Jii 266 INTEGRATION where f(x) x is the derivative of < (x). We have therefore m A where A = m = F(x?) F(xJ and/(z) = -j-F (x). See 171. This is the abscissa of the centroid. To find the ordinate we may use the result of the first example and write for the moment of any element dA = y dx about OX (9) dM=|.<M=|. Whe (10) 2 ' 2 2 Whence (f(aO) 2 where is the derivative of 6 (x~). Therefore (11) where A = m as before. 1. Find the coordinates of the centroid of the area under y 2 = 12 x, from x = 2 to x = 8. 2. Find the coordinates of the centroid of the area of the right triangle whose vertices are (0, 10), (16, 0), (0, 0). Note. This is to be considered as the area under the hypotenuse. Hence find equation of hypotenuse and proceed as above. 3. Find the coordinates of the centroid of the area of the rectangle whose vertices are (1, 0), (6, 0), (1, 7), (6, 7). Note. Consider this area as lying under the upper side. 4. Find the distance from the vertex to the centroid of the triangle whose altitude is h and base 6. Note. . Let the origin be the vertex and the base parallel to the y-axis. Take I as the length of any element. Elimi- INTEGRATION BY PARTS 267 nate I in terms of x by use of similar triangles which will natu- rally occur in the diagram of the problem. 5. Find the distance from the vertex to the centroid of the cone of revolution formed by revolving about OX the right triangle whose vertices are (0, 0), (10, 0), (10, 6). Note. The elements are now circular cylinders of altitude dx and radius y, where y is the ordinate of any point on the hypotenuse of the triangle. 6. Find the abscissa of the centroid of the arc of length I and radius r. Note. Draw the arc so the origin is the center and the x-axis bisects the arc. The element of arc is da, its abscissa is r cos 6, where = a/r in radians measured from OX. Take limits from ~ to ~ or for 6, -^- to =- *9 *9 s Y / 7. Find the centroid of a thin rod whose density varies as its distance from the left end, the density being 5 at unit dis- tance, the rod being 16 units long. Note. The element of mass is now 5 x dx. 8. Solve Ex. 4 above if the density varies as the distance from the vertex and is 3.5 at unit distance. 9. Solve Ex. 5, under same conditions as Ex. 8. 177. Integration by parts. One of the most useful meth- ods of integration when no formula applies directly is inte- gration by parts. It depends on the possibility of separating the given integrand into two factors one of which can be inte- grated directly. Consider (1) 7! (uv) = d/dx (uv) dx = udv vdu; whence (2) I udv = uv I v du, where dv is the integrable factor mentioned above. To apply this formula consider r x sin x dx. 268 INTEGRATION Take 'x = u, sin x dx = dv and substitute in the formula I x sin x dx = x ( cos x) I cos x dx = x cos x + sin x + C. In this example take sin x = u and xdx = dv. Then we obtain by substituting in the formula C x 2 . Cx* I x sm x dx = -~ sin x I -_- cos x dx. The last integrand is more complicated than the original. These results teach us that the choice of factors of the inte- grand is a matter of vital importance. Sometimes only re- peated trials will reveal which set of factors will lead to an integration. Integrate the following: 1. x log xdx. 4. xe ax dx. 2. (sec 2 x 1) x dx. 5. log x dx. 3. x cos xdx. 6. xe x dx. 178. From 171 it is easily seen that if the arc AB is small the chord AB is nearly equal to it. This idea will enable us to employ integration to determine the length of an arc of a curve when its equation is given. For as in geometry we regard the circle as the limit of the sum of the sides of an in- scribed polygon so we now regard the curve PQ as the limit of the sum of such chords as AB. Hence if we write (1) chord AB = VAC 2 + BC 2 = y 1 + f^ AC, and put AC = Ax, BC = Ay&ndAB = As we obtain, noting that A* - o Ax dx (2) arc PQ = Sometimes it may be desirable to use the form (3) > arcP INTEGRATION 269 which is easily seen to be equivalent to the above formula, the difference being that we now integrate with respect to y instead of x and must use y-limits. 1. Find the length of an arc of y* = x 3 from the origin to the point (4, 8). From the given equation y = x%> ^/ 3 t dx~2 X ' arc = Ida -if \/4 + 9xdx = f*(4 + 9x)*dx Jo Jo Jo 2. Find, the length of the catenary y = & e * from x = to x = 10. 3. Find the length of y = sin x from x = to x = IT. 4. Find the length of x$ + y$ = a from z = tojc = a. ,, c , , ,, , . , fx = a0 asinfl. 5. Find the length of one arch of the cycloid < [y = a a cos 6. Note. Remember-^ = , y , ... Use some formulas from 64 dx dx/dd to reduce the integrand to simpler form. 179. Very often an integrand may be simplified and made to fit some formula of integration by a transformation of the variable (see Chap. XIII) or what is ordinarily called a sub- /x dx . Let x = z 2 . Then 1 + x* dx = 2 z dz. Substituting in the given integrand 2 + I = - * 2 + 22 - 21og (z + 1) + C. Restoring x this becomes lx$ ; - x -f 2 a;* - \ log (x* + 1) + C. 270 INTEGRATION Consider another example dx Let x = tan 2. Then dx = sec 2 z dz and Making the substitution there results .= I seczdz J /(sec 2 + tan z), r sec 2 tan 2 + sec 2 2 , sec 2) (dz = I dz (sec z + tan 2) J sec 2 + tan z i*d(secz + tan 2) sec 2 + tan 2 Restoring x we obtain dx = log (sec 2 + tan 2) + C. Vl + z 2 Success with the method of substitutions depends upon wide experience. The number of substitution relations is unlim- ited. The student should consult larger texts on calculus for further treatment. 1. Integrate I , - by the substitution x = - J V(i + x 2 ) 3 y r x \ - i 2. Integrate I -j dx by the substitution x = z 4 . J x 2 + 1 /dx . by the substitution 1 + bx = 2 2 . VI + bx On p. 276 is a more extensive table of integration formulas than is given in 168. The purpose of a table of integrals such as this is to save the time of the student and the practic- ing mathematician after he has sufficient exercise in the proc- ess of integration to ensure that he thoroughly understands the significance of the formulas that he is using. The table may be used in solving the following problems. For a more extensive table of integrals the student is referred to Hudson and Lipka's Table of Integrals. SUPPLEMENTARY EXERCISES 271 ADDITIONAL DERIVATIVES du 1. -7- (arc sin w) = 2. -T- (arc tan w) = 3. -r- (arc sec w) = d , x = -r- (arc cos w). da: v = -T- (arc tan u). dw d , N da: 4. -j- (arc vers u) = . da: \/2 u - tt / = j- (arc esc da: d , . = j- (arc covers u). 5. From vers w = 1 cos u find -T- (vers M) = sin w -^ . da: da: 6. From covers u = 1 sin u find 3- (covers w) = cos u-r- dx da; SUPPLEMENTARY EXERCISES Find the derivatives of the following: * n **./~9 ; i tan x 2. x 3 v a 2 x 2 H 1. x* + e2 cos x. 3. or* x sec x H , Vl -x* 5. Log (1 x 2 ) log sin x 2 . 4. e~z s (1 ax). 6. 3 sin 3 x a cos nx. a + 6 sin x 10. sin x cos 3 x. 12. <&** (1 - x 2 ). 14. log(l - 3X + 4X 2 ). 11. x 4 sin 5 x. 1 l8 x 13> x 2 16 loe 1 "*. I0g 3 + x 17. arc tan ox*. 19. sinx*. 21. arc sec x 18. x 4 arc cos x. 20. e arcsinx . 22. 6 sin e~&. 272 Integrate the following: 1. f l x*dx. ^ dx INTEGRATION vT 7. l sin x cos x dx. Jo 9 f**?. ' J e* 11. j arc sin x do;. 13. fx 2 logxdx. 15. f tan 4 ox dx. > fa tan 5 x sec 3 x dx dx. 21. f sin 2 xcos 2 xdx. 23. J2i 25. foe 5dx /2 4. x sin x dx. Jo 6. 8. | sin 2 x dx. 10. fxe**dx. 12. J*xtan 3 xdx. 14. Jx 2 arcsecxdx. 16. Jcot 3 xcscxdo;. ' J 1 +x 2 ' IT /2 20. I sin 3 x cos 3 x dx. /o o(x a) 3 , 22. 24. 26. / VlOx-25x 2 SUPPLEMENTARY PROBLEMS 1. A train moves out of station A on a straight track. Its distance s in feet from the station is given at any time, until it reached its maximum velocity, by the equation of motion s = 80 P + 30 I, where t is the time in minutes since the train left the station. Find how far the train travels during the first 3 minutes; the first 2 minutes. Find the average velocity of the train during the period of time: (o) t = 2 until t = 3. (6) t = 2 until t = 2.5. (c) t = 2untiH = 2.1. (d) t = 2 until t = 2.001. SUPPLEMENTARY PROBLEMS 273 Find the exact velocity of the train when t = 2 and compare it with the average velocities obtained by the above arithmetical method. 2. The displacement, s, of a body is given by the equation s = 5 + 2 1 2 + t 3 , where s is expressed in feet and t in seconds. Find its average acceleration for 0.001 sec., after the instant t = 2 and compare it with the exact value of the acceleration at the beginning of the interval. 3. If s = i gt 2 , where g = 32.16 ft. /sec 2 and t is expressed in seconds, find the average velocity for 0.01 after the instant t = 3 and compare it with the exact value of the velocity when t = 3. 4. If a body moves so that its horizontal and its vertical distances from the starting point are, respectively, x = 16 t 2 and y = 4 t, show that the equation of its path is y 2 = x and that its horizontal velocity and vertical velocity are, respectively, 32 T and 4 at the instant t = T. 5. If a ball is constrained to move down a plane inclined at an angle 6 with the horizontal, the equation of motion is s = -| (g sin 0) t 2 . Find that value of 6 that will give a maximum velocity for a given value of t. 6. The strength of a rectangular beam varies as the breadth and the square of the depth. Find the dimensions of the strongest beam that can be cut from the log whose diameter is 2 a. 7. In measuring an electric current by means of a tangent galvanometer the percentage error due to a given small error in the reading is proportional to tan x + . Show that this is a minimum when x = 45. tan x 8. The force exerted by a circular electric current of radius a on a small magnet whose axis coincides with the axis of the circle varies as ; > (a 2 + z 2 ) 5 where x = distance of the magnet from the plane of the circle. Prove that the force is a maximum when x = a/2. 9. Find the maximum parallelepiped that can be cut from a sphere if one side of the base is twice the other. 10. Assuming that the current in a voltaic cell is C = ^' where E T -\- K and r are constants representing electromotive force and internal resistance respectively, and R is the external resistance, and that the power given out is P = RC 2 ; show that, if different values are given to R, P will be a maxi- mum when R '= r. 11. A box is to be made from a square piece of cardboard a inches on a side by cutting out squares from the comers and turning up the sides to form the box. Find the side of the square cut out in order that the volume of the box may be a maximum. 12. A water tank 20 ft. high stands on the top of a scaffolding 125 ft. high. At what distance from the base of the scaffolding should one stand in order that the height of the tank might subtend the largest angle at the eye? 274 INTEGRATION 13. If there are n voltaic cells each having an electromotive force of e volts and internal resistance of r ohms, and if x cells are arranged in series and n/x rows in parallel, the current that the battery will send through an external resistance R is given by n xe C = amperes. +R n If n = 20, e = 1.9 volts, r = 0.2 ohm, R = 0.25 ohm; how many cells must be in series to give the greatest possible current? 14. If a body moves so that its horizontal and vertical distances from a point are, respectively, x = 10 t, y = 16 P + 10 t, find its horizontal 16 x 2 speed and its vertical speed. Show that the path is y = + x and 1UU that the slope of the path is the ratio of the vertical speed to the horizontal speed. 15. A point describing the circle x 2 + y 2 = 25 passes through (3, 4) with a velocity of 20 ft. per second. Find its component velocities parallel to the axes. 16. A body moves according to the law s = cos (nt + e). Show that its acceleration is proportional to the space through which it has moved. 17. Find the expression for acceleration for the motion described by the equation x = e at (ci cos bt + Cz sin 6f). 18. If a body is heated to a temperature 61 and then allowed to cool by radiation, its temperature at the time t seconds is given by the equation = Qtf at, where a is a constant. Prove that the rate of cooling is proportional to the temperature. 19. If a point referred to rectangular coordinates moves according to the law x = a cos t -{- b and y = o sin t + c, show that its velocity has a constant magnitude. 20. If a point moves according to the law s = gf* + vd* + SD find velocity as a function of t, the acceleration as a function of t, and the velocity as a function of s. 21. For a beam carrying a uniformly distributed load, w, per unit length, and fixed at one end, ^ - (1 - XV dx*~ 2EI (i Find y in terms of x. 22. Find expressions for velocity and distance when the acceleration is given by a = m nfc 2 cos kt. Determine the constants of integration, Ci and C 2 , by taking v = and s = when t = 0. SUPPLEMENTARY PROBLEMS 275 23. In a chemical reaction of the first order, where a is the initial con- centration of a substance and x is the amount of substance transformed in a time 1, the velocity of the reaction is given by the formula dx/dt = k(a x). Express k as a function of t and x, and x as a function of k and l. 24. A point has an acceleration expressed by the equation a = no 2 cos wt, where r and w are constants. Derive expressions for the velocity and the distance traveled if s = r and v = when t = 0. 25. If y is the deflection at distance x from the fixed end of uniform beam of length I, fixed at one end and loaded with a weight w at the other, then, if we neglect the weight of the beam, J*_fl-aa*L dx* U X) El E and / are constants depending upon the material and shape of the beam. It is known that the deflection y and slope dy/dx are at the fixed end where x = 0. Find an expression for y in terms of x. 26. If the electromotive force, E.M.F., of an alternating current is represented by a sin curve, any ordinate represents the E.M.F. at that point. Show that E a = 0.637 E, where E = maximum E.M.F. and E a = average E.M.F. 27. Another value of importance in the treatment of alternating cur- rents is the square root of the mean square of the ordinates, or the effective E.M.F. Show that E e = 0.707 E, where E e = the effective E.M.F. 28. Air expands isothermally (without change of temperature) accord- ing to Boyle's law, pv = c, where c is a constant. The work done by such an expansion while the volume changes from Vi to v 2 may be represented by the area under the curve p = c/v and between the ordinates v = Vi and v = v 2 . Sketch the curve and determine this area. Find the work done if the expansion continues indefinitely, that is, if v oo . Give a physical interpretation of your result. 29. The equation representing the adiabatic expansion of a gas is pt>* = c, where c is a constant. Find the work done by such an expansion by finding the area under the curve p c/v k and between the ordinates v = Vi and v = v z . Find the work done if v 2 > oo . Give a physical interpretation of your result. Note, During the adiabatic expansion of a gas heat is not communi- cated to nor abstracted from the gas. 30. Derive the four general equations of motion, having given that s=0 and v = VD when t = 0. That is, find v as function of t, s as a func- tion of t, s as a function of t; and derive the formula s = \ (vo + v) t. 276 INTEGRATION TABLE OF INTEGRALS C u n + l 1. (a) l u du = i + C. n^ J n+l u 2 - <> (6) (c) I e nx dx = -e nx + C. J n 3. (a) I udv = uv I v du. (6) / ze* da; = e z (a; - 1) + C. (c) / x^ 1 dz = e*(x* - 2 x + 2) + C. (d) I logxdx = x log # x + C. 4 - t- C du . u . ~ u , o. / = arc sin - + C or arc cos - + C. J Va 2 w 2 a ._ u , = -arc sec - + C or -- arc esc - + C. / - - -- - V w 2 a 2 o a a a du 1 . u a du 1 , a + TABLE OF INTEGRALS 277 C du - 1 l og M ~ a I C J (u - a) (u - 6) a - b g w - 6 ^ 8- a 2 (6) f J . V(w - a) (6 - u) 6 - a A , V(M a) (M 6) 9. / Va 2 w 2 dw = (w Va 2 w 2 + a 2 arc sin w/a) + C. 10. /Vw 2 a 2 dw = i [w Vw 2 a 2 a 2 log (t* + Vw 2 a 2 )] + C. 11. / sin ax da; = cos ax + C. J a 12. I cos ax dx = - sin ax + C. */ a /I 1 tan oa: dx = - log sec ax + (7 = -- log cos as + C. CL Of 14. I ctn ax dx = - log sin ax + C. / a C Tsec as (tan ax + sec az) dx 15. I sec ax dx = I - 7^ - ; - r - J J (tan ax + sec ax) = - log (sec ax + tan ax) +C = -logtan(j + -jr- a a \4 ^ 1 + sin ax 1 cos (ir/2 + ax) since [(sec ax + tan ax) = - - = = , ,' . - r - = cos ax sm (r/2 + ax) tart (r/2 + ox)]. c C C (ctn ox + csc ox) , 16. I csc axdx = I csc ox 7 - : - ; ( ax J J ( ctn ax + csc ax) = - log (csc ax ctn ax) + C = - log tan -^ + C. a /w 1 sin 2 M dw = ^ T sin 2 w + C. /M 1 cos 2 u du = ^ + T sin 2 w + C. 278 INTEGRATION 19. I sec 2 udu = tan u + C. 20 21 . I tan 2 u du = tan w w + C. . / esc 2 w dw = ctn w + C. 22. I sec w tan u du = sec w + C. 23. / esc w ctn w dw = esc it + C. 24. / . = I esc 2 u du. J sm 2 u J 25. I 5 = I sec 2 w du. J COS 2 M J oc C du (*sec 2 udu , 26. I -r - = I - = log tan u + C. J smwcosw J t&nu 27. I sin 3 u du = I (1 cos 2 w) sin w du. 28. / cos 3 wdw = / (1 sin 2 w) cos w dw. 29. / u sin w du = sin w u cos w + C. 30. I ucosudu = usmu + cos w + (7. /gOX e a * cos bx dx = 2 , 2 (a cos 6x + 6 sin 6x) + C. a ~\ o / e ax s ax sin bx dx = , , (a sin bx b cos bx) + C. a 2 + 6 2 33. / arc sin u du = u arc sin u + Vl w 2 -f- C. 34. / arc cos u du = u arc cos u Vl u z + C. 35. / arc tan u du = u arc tan u \ log (1 + w 2 ) + C. 36. / arc ctn u du = u arc ctn u + $ log (I + w 2 ) -J- C. TABLE OF INTEGRALS 279 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. / ware sin udu = \ [(2 u z 1) arc sin u + u Vl w 2 ]-f C. / ware cos udu = l [(2 w 2 1) arc cos w - u Vl w 2 ] + C. / M arc tan udu = \ [(u 2 + 1) arc tan u u] + C. I w arc ctn udu = \ [(w 2 + 1) arc ctn w + u] + C. C du . u ( . = = arcversm-: (versmw = 1 cosw). J -,n \ n = log (log U) + C. r dM f I - ; - = I J u log w J log w / u n loo- ?/ ^?y ?/ n+1 1 ^ U __ - _ I 4- (7 J U + 1 (n + !) 2 / f , du = = log (u + Vw 2 a 2 ). / Vw 2 a 2 r du = j- Ja 2 -w 2 2a du 2aa- 1 , u a USE OF TABLES The use of tables requires in general a knowledge of inter- polation. The method of interpolation is illustrated in what follows. In making interpolations the corrected result should not contain more significant figures than are given in the table which is used. This often requires the "cutting back" of numbers. For example, in the first illustration below, the product of two differences, 0.055 X 0.67 = 0.03685, was cut back to 0.037 in order that the corrected result would not contain more significant figures than that part of the table from which the corrections were made. Note that the last figure retained in the correction was raised one unit. When the part cut off is equal to or greater than one-half the next higher unit, then that unit is increased by one if less than one-half no change is made in that unit. TABLE I: Powers and roots of numbers from i to 100. Assuming that the roots of successive numbers are propor- tional to their corresponding numbers, we may find the root of a number that does not appear in the table, such as 83.67, by interpolation as follows : From the table V84 = 9.165. V83 = 9.110. A difference of 1 in the numbers causes a difference of 0.055 in their roots. Therefore, a difference of 0.67 in the numbers causes a difference of 0.037 ( = 0.67 X 0.055) in the roots. Therefore, V83.67 = 9.110 + 0.037 = 9.147. The root of a decimal fraction such as the cube root of 0.06831 may be obtained from the table as follows: USE OF TABLES 281 From the table ^69 = 4.102. ^68 = 4.082. A difference of 1 in the numbers causes a difference of 0.02 in their roots. Therefore, a difference of 0.31 in the numbers causes a difference of 0.006 in the roots. Therefore, ^68.31 - 4.082 + 0.006 = 4.088 and ^0.06831 = ^ ^68.31 = 0.4088. In this illustration notice that the number was first multi- plied by 1000 and the cube root of this new number was found from the table. This root when divided by 10 gave the root sought. To find the square root of the decimal fraction, 0.06831, multiply the number by 100. Then find the root of this num- ber (6.831) and divide this root (2.613) by 10 and obtain 0.2613 as the square root of 0.06831. TABLE II: How to find the logarithm of a number. Find log 27.4. The characteristic by rule is 1. To obtain the mantissa from the table, look for 27 in the column headed N. Opposite 27 and in the column headed by 4 find the number 4378, which is the mantissa with decimal point omitted. Therefore, log 27.4 = 1.4378. Find log 274.3. The mantissa of this logarithm is not given directly in the table. If we assume that a small change in the number causes a proportional change in the logarithm, then we may proceed by interpolation as follows: mantissa of log 275 is 0.4393. mantissa of log 274 is 0.4378. We observe that a difference of 1 in the number makes a difference of 0.0015 in the logarithm, or a difference of 0.3 in the number makes a difference of 0.0015 X 0.3 = 0.00045, or cutting the number back one place, our correction is 0.0005. The logarithm of 274.3 = 2.4378 + 0.0005 = 2.4383. Time will be saved in interpolating by considering the mantissas for 282 USE OF TABLES the moment as whole numbers. Thus, instead of writing 0.0015, write 15, and so forth. How to find the number corresponding to a given loga- rithm. Given log N = 2.4383. To find the number N. Since this problem is the converse of the preceding one, we may trace that problem back. We cannot find the mantissa 0.4383 in the table, but we find 0.4393 and 0.4378 and so forth. Mechanical interpolations. In order to facilitate the com- putation, the tabular difference and the proportional part for the fourth figure of the natural numbers is given at the bot- tom of the page. The student is advised not to use this part of the table until he has learned to interpolate mentally with speed and accuracy. In scientific work one is called upon to use many different tables in which tabular differences and pro- portional parts are not given. For this reason, the student should learn to be independent of these aids. In the above problems beginning with the tabular difference 4393 - 4378 = 15, look at bottom of page under "Tab. Difif." for 15 which is found on the third page of the tables. Opposite 15 and in column headed 3 find 4.5. Add this correction, after cutting it back according to rule, to the mantissa 4378 and obtain 4383, the same mantissa as above with decimal point omitted. If log N = 2.4383, find the number N, proceeding as follows: Find mantissa in the table nearest less than 4383. It is 4378, which is in column headed 4 and opposite the number 27 in the first column. Hence 0.4378 is the mantissa of the loga- rithm of 274. Now 4383-4378= 5. 4393 - 4378 = 15. At bottom of page under "Tab. Diff." opposite 15, find the number nearest 5. It is 4.5 in column headed 3. Therefore, 3 is the next figure (fourth) of the number sought. The deci- mal point, by rule for characteristic, should be placed so that the integral part of the number will have three digits. There- fore, the number is 274.3. USE OF TABLES 283 Use of Tables of Trigonometric Functions In Table III are given the natural and logarithmic functions of angles for every 10 minutes in the quadrant. The angles less than 45 are found in the left-hand column. The func- tions are given in same line with the angle. Angles from 45 to 90 are found in the right-hand column. It should be noted that when angles are read on left, function names are to be read at top of page and when angles are read in right column, function names are to be read at bottom of page. To find the logarithmic sine of 41 20.' Since this angle is less than 45 read the angle in left column. On the line of 41 20' in column headed log sine find log sin 41 20' = 9.8198. The table is so constructed that the logarithmic functions are 10 larger than their actual values. This will be understood if the logarithm of the natural sine of 41 20' is found and its logarithm found in Table II. The adding of 10 to the loga- rithmic functions is a matter of facility in calculating with them. To find an angle corresponding to a given function we proceed in a manner similar to finding a number when its logarithm is given. 284 TABLES I. POWERS AND ROOTS OF NUMBERS FROM 1 TO 100 No. Square. Cube. Square root. Cube root. 1 1 1 1.000 1.000 2 4 8 1.414 1.260 3 9 27 1.732 1.442 4 16 64 2.000 1.587 5 25 125 2.236 1.710 6 36 216 2.450 1.817 7 49 343 2.646 1.913 8 64 512 2.828 2.000 9 81 729 3.000 2.080 10 100 1000 3.162 2.154 11 121 1331 3.317 2.224 12 144 1728 3.464 2.289 13 169 2197 3.606 2.351 14 196 2744 3.742 2.410 15 225 3375 3.873 2.466 16 256 4096 4.000 2.520 17 289 4913 4.123 2.571 18 324 5832 4.243' 2.621 19 361 6859 4.359 2.668 20 400 8000 4.472 2.714 21 441 9261 4.583 2.759 22 484 10648 4.690 2.802 23 529 12167 4.796 2.844 24 576 13824 4.899 2.885 25 625 15625 5.000 2.924 26 676 17576 5.099 2.963 27 729 19683 5.196 3.000 28 784 21952 5.292 3.037 t 29 841 24389 5.385 3.072 30 900 27000 5.477 3.107 31 961 29791 5.568 3.141 32 1024 32768 5.657 3.175 33 1089 35937 5.745 3.208 34 1156 39304 5.831 3.240 35 1225 42875 5.916 3.271 36 1296 46656 6.000 3.302 37 1369 50653 6.083 3.332 38 1444 54872 6.164 3.362 39 1521 59319 6.245 3.391 40 1600 64000 6.325 3.420 41 1681 68921 6.403 3.448 42 1764 74088 6.481 3.476 43 1849 79507 6.557 3.503 44 1936 85184 6.633 3.530 45 2025 91125 6.708 3.557 46 2116 97336 6.782 3.583 47 2209 103823 6.856 3.609 48 2304 110592 6.928 3.634 49 2401 117649 7.000 3.659 50 2500 125000 7.071 3.684 TABLES 285 I. POWERS AND ROOTS OF NUMBERS FROM 1 TO 100. (Cont'd) No. Square. Cube. Square root. Cube root. 51 2601 132651 7.141 3.708 52 2704 140608 7.211 3.733 53 2809 148877 7.280 3.756 54 2916 157464 7.349 3.780 55 3025 166375 7.416 3.803 56 3136 175616 7.483 3.826 57 3249 185193 7.550 3.849 58 3364 195112 7.616 3.871 59 3481 205379 7.681 3 893 60 3600 216000 7.746 3.915 61 3721 226981 7.810 3.937 62 3844 238328 7.874 3.958 63 3969 250047 7.937 3.979 64 4096 262144 8.000 4.000 65 4225 274625 8.062 4.021 66 4356 287496 8.124 4.041 67 4489 300763 8.185 4.062 68 4624 314432 8.246 4.082 69 4761 328509 8.306 4.102 70 4900 343000 8.367 4.121 71 5041 357911 8.426 4.141 72 5184 373248 8.485 4.160 73 5329 389017 8.544 4.179 74 5476 405224 8.602 4.198 75 5625 421875 8.660 4.217 76 5776 438976 8.718 4.236 77 5929 456533 8.775 4.254 78 6084 474552 8.832 4.273 79 6241 493039 8.888 4.291 80 6400 512000 8.944 4.309 81 6561 531441 9.000 4.327 82 6724 551368 9.055 4.345 83 6889 571787 9.110 4.362 84 7056 592704 9.165 4.380 85 7225 614125 9.220 4.397 86 7396 636056 9.274 4.414 87 7569 658503 9.327 4.431 88 7744 681472 9.381 4.448 89 7921 704969 9.434 4.465 90 8100 729000 9.487 4.481 91 8281 753571 9.539 4.498 92 8464 778688 9.592 4.514 93 8649 804357 9.644 4.531 94 8836 830584 9.695 4.547 96 9025 857375 9.747 4.563 96 9216 884736 9.798 4.579 97 9409 912673 9.849 4.595 98 9604 941192 9.900 4.610 99 9801 970299 9.950 4.626 100 10000 1000000 10.000 4.642 286 TABLES II. LOGARITHMS No. 1 2 3 4 5 6 7 8 9 Io~ 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 Tab. diff. Extra digit. 1 2 3 4 5 6 7 8 9 43 4.3 8.6 12.9 17.2 21.5 25.8 30.1 34.4 38.7 42 4.2 8.4 12.6 16.8 21.0 25.2 29.4 33.6 37.8 41 4.1 8.2 12.3 16.4 20.5 24.6 28.7 32.8 36.9 40 4.0 8.0 12.0 16.0 20.0 24.0 28.0 32.0 36.0 39 3.9 7.8 11.7 15.6 19.5 23.4 27.3 31.2 35.1 38 3.8 7.6 11.4 15.2 19.0 22.8 26.6 30.4 34.2 37 3.7 7.4 11.1 14.8 18.5 22.2 25.9 29.6 33.3 36 3.6 7.2 10.8 14.4 18.0 27.6 25.2 28.8 32.4 35 3.5 7.0 10.5 14.0 17.5 21.0 24.5 28.0 31.5 34 3.4 6.8 10.2 13.6 17.0 20.4 23.8 27.2 30.6 33 3.3 6.6 9.9 13.2 16.5 19.8 23.1 26.4 29.7 32 3.2 6.4 9.6 12.8 16.0 19.2 22.4 25.6 28.8 31 3.1 6.2 9.3 12.4 15.5 18.6 21.7 24.8 27.9 30 3.0 6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 TABLES 287 II. LOGARITHMS (Continued) No. l 2 3 4 5 6 7 8 9 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 61 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241' 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 Tab. diff. Extra digit. 1 2 3 4 5 6 7 8 9 29 2.9 5.8 8.7 11.6 14.5 17.4 20.3 23.2 26.1 28 2.8 5.6 8.4 11.2 14.0 16.8 19.6 22.4 25.2 27 2.7 5.4 8.1 10.8 13.5 16.2 18.9 21.6 24.3 26 2.6 5.2 7.8 10.4 13.0 15.6 18.2 20.8 33.4 25 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 24 2.4 4.8 7.2 9.6 12.0 14.4 16.8 19.2 21.6 23 2.3 4.6 6.9 9.2 11.5 13.8 16.1 18.4 20.7 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 20 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 19 1.9 3.8 5.7 7.6 9.5 11.4 13.3 15.2 17.1 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 288 TABLES II. LOGARITHMS (Continued) No. 1 2 3 4 5 6 7 8 9 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 Tab. Diff. Extra digit. 1 2 3 4 5 6 7 8 9 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 14 1.4 2.8 4.2 5.6 7.0 8.4 9.8 11.2 12.6 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 12 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 10 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 9 0.9 1.8 2.7 3.6 4.5 5.4 6.3 7.2 8.1 8 0.8 1.6 2.4 3.2 4.0 4.8 5.6 6.4 7.2 7 0.7 1.4 2.1 2.8 3.5 4.2 4.9 5.6 6.3 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4 5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 4 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 TABLES 2S9 III. NATURAL AND LOGARITHMIC FUNCTIONS Sine. Cosine. Tangent. Cotangent. Angle. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 0.0000 00 1.0000 0.0000 0.0000 00 00 00 90 10' .0029 7.4637 1.0000 .0000 .0029 7.4637 343.8 12.5363 50' 20' .0058 .7648 1.0000 .0000 .0058 .7648 171.9 .2352 40' 30' .0087 .9408 1.0000 .0000 .0087 .9409 114.6 .0591 30' 40' .0116 8.0658 0.9999 .0000 .0116 8.0658 85.94 11.9342 20' 50' .0145 .1627 .9999 .0000 .0145 .1627 68.75 .8373 10' 1 .0175 .2419 .9998 9.9999 .0175 .2419 57.29 .7581 89 10' .0204 .3088 .9998 .9999 .0204 .3089 49.10 .6911 50' 20' .0233 .3668 .9997 .9999 .0233 .3669 42.96 .6331 40' 30' .0262 .4179 .9997 .9999 .0262 .4181 38.19 .5819 30' 40' .0291 .4637 .9996 .9998 .0291 .4638 34.37 .5362 20' 50' .0320 .5050 .9995 .9998 .0320 .5053 31.24 .4947 10' 2 .0349 .5428 .9994 .9997 .0349 .5431 28.64 .4569 88 10' .0378 .5776 .9993 .9997 .0378 .5779 26.43 .4221 50' 1 20' .0407 .6097 .9992 .9996 .0407 .6101 24.54 .3899 40' ft 30' .0436 .6397 .9990 .9996 .0437 .6401 22.90 .3599 30' 1 40' .0465 .6677 .9989 .9995 .0466 .6682 21.47 .3318 20' 50' .0494 .6940 .9988 .9995 .0495 .6945 20.21 .3055 10' * 3 .0523 .7188 .9986 .9994 .0524 .7194 19.08 .2806 87 10' .0552 .7423 .9985 .9993 .0553 .7429 18.08 .2571 50' 20' .0581 .7645 .9983 .9993 .0582 .7652 17.17 .2348 40' 30' .0610 .7857 .9981 .9992 .0612 .7865 16.35 .2135 30' 40' .0640 .8059 .9980 .9991 .0641 .8067 15.61 .1933 20' 50' .0669 .8251 .9978 .9990 .0670 .8261 14.92 .1739 10' 4 .0698 .8436 .9976 .9989 .0699 .8446 14.30 .1554 86 10' .0727 .8613 .9974 .9989 .0729 .8624 13.73 .1376 50' 20' .0756 .8783 .9971 .9988 .0758 .8795 13.20 .1205 40' 30' .0785 .8946 .9969 .9987 .0787 .8960 12.71 .1040 30' 40' .0814 .9104 .9967 .9986 .0816 .9118 12.25 .0882 20' 50' .0843 .9256 .9964 .9985 .0846 .9272 11.83 .0728 10' 5 .0872 .9403 .9962 .9983 .0875 .9420 11.43 .0580 85 10' .0901 .9545 .9959 .9982 .0904 .9563 11.06 .0437 50' 20' .0929 .9682 .9957 .9981 .0934 .9701 10.71 .0299 40' 30' .0958 .9816 .9954 .9980 .0963 .9836 10.39 .0164 30' 40' .0987 .9945 .9951 .9979 .0992 .9966 10.08 .0034 20' 50' .1016 9.0070 .9948 .9977 .1022 9.0093 9.788 10.9907 10' Nat. Log. Nat. Log. Nat. Log. Nat. Log. A n ~1 A Cosine. Sine. Cotangent. Tangent. Ann !e. For angles over 45 use right column and take names of functions at bottom of page. 290 TABLES III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent. Angle. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 6 0.1045 9.0192 0.9945 9.9976 0.1051 9.0126 9.514 10.9784 84 10' .1074 .0311 .9942 .9975 .1080 .0336 9.255 .9664 50' 20' .1103 .0426 .9939 .9973 .1110 .0453 9.010 .9547 40' 30' .1132 .0539 .9936 .9972 .1139 .0567 8.777 .9433 30' 40' .1161 .0648 .9932 .9971 .1169 .0678 8.556 .9322 20' 50' .1190 .0755 .9929 .9969 .1198 .0786 8.345 .9214 10' 7 .1219 .0859 .9925 .9968 .1228 .0891 8.144 .9109 83 10' .1248 .0961 .9922 .9966 .1257 .0995 7.953 .9005 50' 20' .1276 .1060 .9918 .9964 .1287 .1096 7.770 .8904 40' 30' .1305 .1157 .9914 .9963 .1317 .1194 7.596 .8806 30' 40' .1334 .1252 .9911 .9961 .1346 .1291 7.429 .8709 20' 50' .1363 .1345 .9907 .9959 .1376 .1385 7.269 .8615 10' 8 .1392 .1436 .9903 .9958 .1405 .1478 7.115 .8522 82 10' .1421 .1525 .9899 .9956 .1435 .1569 6.968 .8431 50' I 20' .1449 .1612 .9894 .9954 .1465 .1658 6.827 .8342 40' -8 30' .1478 .1697 .9890 .9952 .1495 .1745 6.691 .8255 30' -g 8 40' .1507 .1781 .9886 .9950 .1524 .1831 6.561 .8169 20' S 50' .1536 .1863 .9881 .9948 .1554 .1915 6.435 .8085 10' 9 .1564 .1943 .9877 .9946 .1584 .1997 6.314 .8003 81 10' .1593 .2022 .9872 .9944 .1614 .2078 6.197 .7922 50' 20' .1622 .2100 .9868 .9942 .1644 .2158 6.084 .7842 40' 30' .1650 .2176 .9863 .9940 .1673 .2236 5.976 .7764 30' 40' .1679 .2251 .9858 .9938 .1703 .2313 5.871 .7687 20' 50' .1708 .2324 .9853 .9936 .1733 .2389 5.769 .7611 10' 10 .1736 .2397 .9848 .9934 .1763 .2463 5.671 .7537 80 10' .1765 .2468 .9843 .9931 .1793 .2536 5.576 .7464 50' 20' .1794 .2538 .9838 .9929 .1823 .2609 5.485 .7391 40' 30' .1822 .2606 .9833 .9927 .1853 .2680 5.396 .7320 30' 40' .1851 .2674 .9827 .9924 .1883 .2750 5.309 .7250 20' 50' .1880 .2740 .9822 .9922 .1914 .2819 5.226 .7181 10' 11 .1908 .2806 .9816 .9919 .1944 .2887 5.145 .7113 79 10' .1937 .2870 .9811 .9917 .1974 .2953 5.066 .7047 50' 20' .1965 .2934 .9805 .9914 .2004 .3020 4.989 .6980 40' 30' .1994 .2997 .9799 .9912 .2035 .3085 4.915 .6915 30' 40' .2022 .3058 .9793 .9909 .2065 .3149 4.843 .6851 20' 50' .2051 .3119 .9787 .9907 .2095 .3212 4.773 .6788 10' Nat. Log. Nat. Log. Nat. Log. Nat. Log. A (in-lst Cosine. Sine. Cotangent. Tangent. Angle. For angles over 45 use right column and take names of functions at bottom of page. TABLES 291 III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent. A 1 Nat. Log. , Nat. Log. Nat. Log. Nat. Log. 12 0.2079 9.3179 0.9781 9.9904 0.2126 9.3275 4.705 10.6725 78 10' .2108 .3238 .9775 .9901 .2156 .3336 4.638 .6664 50' 20' .2136 .3296 .9769 .9899 .2186 .3397 4.574 .6603 40' 30' .2164 .3353 .9763 .9896 .2217 .3458 4.511 .6542 30' 40' .2193 .3410 .9757 .9893 .2247 .3517 4.449 .6483 20' 50' .2221 .3466 .9750 .9890 .2278 .3576 4.390 .6424 10' 13 .2250 .3521 .9744 .9887 .2309 .3634 4.332 .6366 77 10' .2278 .3575 .9737 .9884 .2339 .3691 4.275 .6309 50' 20'~ .2306 .3629 .9730 .9881 .2370 .3748 4.219 .6252 40' 30' .2334 .3682 .9724 .9878 .2401 .3804 4.165 .6196 30' 40' .2363 .3734 .9717 .9875 .2432 .3859 4.113 .6141 20' 50' .2391 .3786 .9710 .9872 .2462 .3914 4.061 .6086 10' 14 .2419 .3837 .9703 .9869 .2493 .3968 4.011 .6032 76 10' .2447 .3887 .9696 .9866 .2524 .4021 3.962 .5979 50' 1 20' .2476 .3937 .9689 .9863 .2555 .4074 3.914 .5926 40' 30' .2504 .3986 .9681 .9859 .2586 .4127 3.867 .5873 30' -3 1 40' .2532 .4035 .9674 .9856 .2617 .4178 3.821 .5822 20' 50' .2560 .4083 .9667 .9853 .2648 .4230 3.776 .5770 10' 15 .2588 .4130 .9659 .9849 .2679 .4281 3.732 .5719 75 10' .2616 .4177 .9652 .9846 .2711 .4331 3.689 .5669 50' 20' .2644 .4223 .9644 .9843 .2742 .4381 3.647 .5619 40' 30' .2672 .4269 .9636 .9839 .2773 .4430 3.606 .5570 30' 40' .2700 .4314 .9628 .9836 .2805 .4479 3.566 .5521 20' 50' .2728 .4359 .9621 .9832 .2836 .4527 3.526 .5473 10' 16 .2756 .4403 .9613 .9828 .2867 .4575 3.487 .5425 74 10' .2784 .4447 .9605 .9825 .2899 .4622 3.450 .5378 50' 20' .2812 .4491 .9596 .9821 .2931 .4669 3.412 .5331 40' 30' .2840 .4533 .9588 .9817 .2962 .4716 3.376 .5284 30' 40' .2868 .4576 .9580 .9814 .2994 .4762 3.340 .5238 20' 50' .2896 .4618 .9572 .9810 .3026 .4808 3.305 .5192 10' 17 .2924 .4659 .9563 .9806 .3057 .4853 3.271 .5147 73 10' .2952 .4700 .9555 .9802 .3089 .4898 3.237 .5102 50' 20' .2979 .4741 .9546 .9798 .3121 .4943 3.204 .5057 40' 30' .3007 .4781 .9537 .9794 .3153 .4987 3.172 .5013 30' 40' .3035 .4821 .9528 .9790 .3185 .5031 3.140 .4969 20' 50' .3062 .4861 .9520 .9786 .3217 .5075 3.108 .4925 10' Nat. Log. Nat. Log. Nat. Log. Nat. Log. Cosine. Sine. Cotangent. Tangent. Angle. For angles over 45 use right column and take names of functions at bottom of page. 292 TABLES III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent. Angle. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 18 0.3090 9.4900 0.9511 9.9782 0.3249 9.5118 3.078 10.4882 72 10' .3118 .4939 .9502 .9778 .3281 .5161 3.048 .4839 50' 20' .3145 .4977 .9492 .9774 .3314 .5203 3.018 .4797 40' 30' .3173 .5015 .9483 .9770 .3346 .5245 2.989 .4755 30' 40' .3201 .5052 .9474 .9765 .3378 .5287 2.960 .4713 20' 50' .3228 .5090 .9465 .9761 .3411 .5329 2.932 .4671 10' 19 .3256 .5126 .9455 .9757 .3443 .537C 2.904 .4630 71 10' .3283 .5163 .9446 .9752 .3476 .5411 2.877 .4589 50' 20' .3311 .5199 .9436 .9748 .3508 .5451 2.850 .4549 40' 30' .3338 .5235 .9426 .9743 .3541 .5491 2.824 .4509 30' 40' .3365 .5270 .9417 .9739 .3574 .5531 2.798 .4469 20' 50' .3393 .5306 .9407 .9734 .3607 .5571 2.773 .4429 10' 20 .3420 .5341 .9397 .9730 .3640 .5611 2.748 .4389 70 10' .3448 .5375 .9387 .9725 .3673 .5650 2.723 .4350 50' I 20' .3475 .5409 .9377 .9721 .3706 .5689 2.699 .4311 40' * I 30' .3502 .5443 .9367 .9716 .3739 .5727 2.675 .4273 30' -g 40' .3529 .5477 .9356 .9711 .3772 .5766 2.651 .4234 20' 50' .3557 .5510 .9346 .9706 .3805 .5804 2.628 .4196 10' 21 .3584 .5543 .9336 .9702 .3839 .5842 2.605 .4158 69 10' .3611 .5576 .9325 .9697 .3872 .5879 2.583 .4121 50' 20' .3638 .5609 .9315 .9692 .3906 .5917 2.561 .4083 40' 30' .3665 .5641 .9304 .9687 .3939 .5954 2.539 .4046 30' 40' .3692 .5673 .9293 .9682 .3973 .5991 2.517 .4009 20' 50' .3719 .5704 .9283 .9677 .4006 .6028 2.496 .3972 10' 22 .3746 .5736 .9272 .9672 .4040 .6064 2.475 .3936 68 10' .3773 .5767 .9261 .9667 .4074 .6100 2.455 .3900 50' 20' .3800 .5798 .9250 .9661 .4108 .6136 2.434 .3864 40' 30' .3827 .5828 .9239 .9656 .4142 .6172 2.414 .3828 30' 40' .3854 .5859 .9228 .9651 .4176 .6208 2.395 .3792 20' 50' .3881 .5889 .9216 .9646 .4210 .6243 2.375 .3757 10' 23 .3907 .5919 .9205 .9640 .4245 .6279 2.356 .3721 67 10' .3934 .5948 .9194 .9635 .4279 .6314 2.337 .3686 50' 20' .3961 .5978 .9182 .9629 .4314 .6348 2.318 .3652 40' 30' .3987 .6007 .9171 .9624 .4348 .6383 2.300 .3617 30' 40' .4014 .6036 .9159 .9618 .4383 .6417 2.282 .3583 20' 50' .4041 .6065 .9147 .9613 .4417 .6452 2.264 .3548 10' Nat. Log. Nat. 1 Log. Nat. Log. Nat. Log A t ,, T |o Consie. Sine. Cotangent. Tangent. /\IlgH?. For angles over 45 use right column and take names of function at bottom of page. TABLES 293 III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent- A n LC 1 c . Nat. Log. Nat. Log. Nat. Log. Nat. Log. 24 0.4067 9.6093 0.9135 9.9607 0.4452 9.6486 2.246 10.3514 66 10' .5094 .6121 .9124 .9602 .4487 .6520 2.229 .3480 50' 20' .4120 .6149 .9112 .9596 .4522 .6553 2.211 .3447 40' 30' .4147 .6177 .9100 .9590 .4557 .6587 2.194 .3413 30' 40' .4173 .6205 .9088 .9584 .4592 .6620 2.178 .3380 20' 50' .4200 .6232 .9075 .9579 .4628 .6654 2.161 .3346 Iff 25 .4226 .6259 .9063 .9573 .4663 .6687 2.145 .3313 65 10' .4253 .6286 .9051 .9567 .4699 .6720 2.128 .3280 50' 20' .4279 .6313 .9038 .9561 .4734 .6752 2.112 .3248 40' 30' .4305 .6340 .9026 .9555 .4770 .6785 2.097 .3215 30' 40' .4331 .6366 .9013 .9549 .4806 .6817 2.081 .3183 20' 50' .4358 .6392 .9001 .9543 .4841 .6850 2.066 .3150 10' 26 .4384 .6418 .8988 .9537 .4877 .6882 2.050 .3118 64 10' .4410 .6444 .8975 .9530 .4913 .6914 2.035 .3086 50' I 20' .4436 .6470 .8962 .9524 .4950 .6946 2.020 .3054 40' R 1 30' .4462 .6495 .8949 .9518 .4986 .6977 2.006 .3023 30' , 1 40' .4488 .6521 .8936 .9512 .5022 .7009 1.991 .2991 20' 1 50' .4514 .6546 .8923 .9505 .5059 .7040 1.977 .2960 10' 27 .4540 .6570 .8910 .9499 .5095 .7072 1.963 .2928 63 10' .4566 .6595 .8897 .9492 .5132 .7103 1.949 .2897 50' 20' .4592 .6620 .8884 .9486 .5169 .7134 1.935 .2866 40' 30' .4617 .6644 .8870 .9479 .5206 .7165 1.921 .2835 30' 40' .4643 .6668 .8857 .9473 .5243 .7196 1.907 .2804 20' 50' .4669 .6692 .8843 .9466 .5280 .7226 1.894 .2774 10' 28 .4695 .6716 .8829 .9459 .5317 .7257 1.881 .2743 62 10' .4720 .6740 .8816 .9453 .5354 .7287 1.868 .2713 50' 20' .4746 .6763 .8802 .9446 .5392 .7317 1.855 .2683 40' 30' .4772 .6787 .8788 .9439 .5430 .7348 1.842 .2652 30' 40' .4797 .6810 .8774 .9432 .5467 .7378 1.829 .2622 20' 50' .4823 .6833 .8760 .9425 .5505 .7408 1.817 .2592 10' 29 .4848 .6856 .8746 .9418 .5543 .7438 1.804 .2562 61 10' .4874 .6878 .8732 .9411 .5581 .7467 1.792 .2533 50' 20' .4899 .6901 .8718 .9404 .5619 .7497 1.780 .2503 40' 30' .4924 .6923 .8704 .9397 .5658 .7526 1.768 .2474 30' 40' .4950 .6946 .8689 .9390 .5696 .7556 1.756 .2444 20' 50' .4975 .6968 .8675 .9383 .5735 .7585 1.744 .2415 10' Nat. Log. Nat. Log. Nat. Log. Nat. Log. A .,?,-, Cosine. Sine. Cotangent. Tangent. Angle. For angles over 45 use right column and take names of functions at bottom of page. 294 TABLES III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent. Angle. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 30 0.5000 9.6990 0.8660 9.9375 0.5774 9.7614 1.732 10.2386 60 10' .5025 .7012 .8646 .9368 .5812 .7644 1.721 .2356 50' 20' .5050 .7033 .8631 .9361 .5851 .7673 1.709 .2327 40' 30' .5075 .7055 .8616 .9353 .5890 .7701 1.698 .2299 30' 40' .5100 .7076 .8601 .9346 .5930 .7730 1.686 .2270 20' 50' .5125 .7097 .8587 .9338 .5969 .7759 1.675 .2241 10' 31 .5150 .7118 .8572 .9331 .6009 .7788 1.664 .2212 59 10' .5175 .7139 .8557 .9323 .6048 .7816 1.653 .2184 50' 20' .5200 .7160 .8542 .9315 .6088 .7845 1.643 .2155 40 7 30' .5225 .7181 .8526 .9308 .6128 .7873 1.632 .2127 30' 40' .5250 .7201 .8511 .9300 .6168 .7902 1.621 .2098 20' 50' .5275 .7222 .8496 .9292 .6208 .7930 1.611 .2070 10' 32 .5299 .7242 .8480 .9284 .6249 .7958 1.600 .2042 58 10' .5324 .7262 .8465 .9276 .6289 .7986 1.590 .2014 50' 20' .5348 .7282 .8450 .9268 .6330 .8014 1.580 .1986 40' a 1 30' .5373 .7302 .8434 .9260 .6371 .8042 1.570 .1958 30' ^ 3 40' .5398 .7322 .8418 .9252 .6412 .8070 1.560 .1930 20' 50' .5422 .7342 .8403 .9244 .6453 .8097 1.550 .1903 10' 33 .5446 .7361 .8387 .9236 .6494 .8125 1.540 .1875 57 10' .5471 .7380 .8371 .9228 .6536 .8153 1.530 .1847 50' 20' .5495 .7400 .8355 .9219 .6577 .8180 1.520 .1820 40' 30' .5519 .7419 .8339 .9211 .6619 .8208 1.511 .1792 30' 40' .5544 .7438 .8323 .9203 .6661 .8235 1.501 .1765 20' 50' .5568 .7457 .8307 .9194 .6703 .8263 1.492 .1737 10' 34 .5592 .7476 .8290 .9186 .6745 .8290 1.483 .1710 56 10' .5616 .7494 .8274 .9177 .6787 .8317 1.473 .1683 50' 20' .5640 .7513 .8258 .9169 .6830 .8344 1.464 .1656 40' 30' .5664 .7531 .8241 .9160 .6873 .8371 1.455 .1629 30' 40' .5688 .7550 .8225 .9151 .6916 .8398 1.446 .1602 20' 50' .5712 .7568 .8208 .9142 .6959 .8425 1.437 .1575 10' 35 .5736 .7586 .8192 .9134 .7002 .8452 1.428 .1548 55 10' .5760 .7604 .8175 .9125 .7046 .8479 1.419 .1521 50' 20' .5783 .7622 .8158 .9116 .7089 .8506 1.411 .1494 40' 30' .5807 .7640 .8141 .9107 .7133 .8533 1.402 .1467 30' 40' .5831 .7657 .8124 .9098 .7177 .8559 1.393 .1441 20' 50' .5854 .7675 .8107 .9089 .7221 .8586 1.385 .1414 10' Nat. Log. Nat. Log. Nat. Log. Nat. Log. A ,1,1,1 Cosine. Sine. Cotangent. Tangent. Angle. For angles over 45 use right column and take names of functions at bottom of page. TABLES 295 III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent. Angle. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 36 0.5878 9.7692 0.8090 9.9080 0.7265 9.8613 1.376 10.1387 54 10' .5901 .7710 .8073 .9070 .7310 .8639 1.368 .1361 50' 20' .5925 .7727 .8056 .9061 .7355 .8666 1.360 .1334 40' 30' .5948 .7744 .8039 .9052 .7400 .8692 1.351 .1308 30' 40' .5972 .7761 .8021 .9042 .7445 .8718 1.343 .1282 20' 50' .5995 .7778 .8004 .9033 .7490 .8745 1.335 .1255 10' 37 .6018 .7795 .7986 .9023 .7536 .8771 1.327 .1229 53 10' .6041 .7811 .7969 .9014 .7581 .8797 1.319 .1203 50' 20' .6065 .7828 .7951 .9004 .7627 .8824 1.311 .1176 40' 30' .6088 .7844 .7934 .8995 .7673 .8850 1.303 .1150 30' 40' .6111 .7861 .7916 .8985 .7720 .8876 1.295 .1124 20' 50' .6134 .7877 .7898 .8975 .7766 .8902 1.288 .1098 10' 38 .6157 .7893 .7880 .8965 .7813 .8928 1.280 .1072 52 10' .6180 .7910 .7862 .8955 .7860 .8954 1.272 .1046 50' 1 20' .6202 .7926 .7844 .8945 .7907 .8980 1.265 .1020 40' a - 30' .6225 .7941 .7826 .8935 .7954 .9006 1.257 .0994 30' 1 40' .6248 .7957 .7808 .8925 .8002 .9032 1.250 .0968 20' 1 50' .6271 .7973 .7790 .8915 .8050 .9058 1.242 .0942 w* 39 .6293 .7989 .7771 .8905 .8098 .9084 1.235 .0916 51 10' .6316 .8004 .7753 .8895 .8146 .9110 1.228 .0890 50' 20' .6338 .8020 .7735 .8884 .8195 .9135 1.220 .0865 40' 30' .6361 .8035 .7716 .8874 .8243 .9161 1.213 .0839 30' 40' .6383 .8050 .7698 .8864 .8292 .9187 1.206 .0813 20' 50' . .6406 .8066 .7679 .8853 .8342 .9212 1.199 .0788 10' 40 .6428 .8081 .7660 .8843 .8391 .9238 1.192 .0762 50 10' .6450 .8096 .7642 .8832 .8441 .9264 1.185 .0736 50' 20' .6472 .8111 .7623 .8821 .8491 .9289 1.178 .0711 40' 30' .6494 .8125 .7604 .8810 .8541 .9315 1.171 .0685 30' 40' .6517 .8140 .7585 .8800 .8591 .9341 1.164 .0659 20' 50' .6539 .8155 .7566 .8789 .8642 .9366 1.157 .0634 10' 41 .6561 .8169 .7547 .8778 .8693 .9392 1.150 .0608 49 10' .6583 .8184 .7528 .8767 .8744 .9417 1.144 .0583 50' 20' .6604 .8198 .7509 .8756 .8796 .9443 1.137 .0557 40' 30' .6626 .8213 .7490 .8745 .8847 .9468 1.130 .0532 30' 40' .6648 .8227 .7470 .8733 .8899 .9494 1.124 .0506 20' 50' .6670 .8241 .7451 .8722 .8952 .9519 1.117 .0481 10' Nat. Log. Nat. Log. Nat. Log. Nat. Log. Cosine. Sine. Cotangent. Tangent. Angle. For angles over 45 use right column and take names of functions at bottom of page. 296 TABLES III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) Sine. Cosine. Tangent. Cotangent. Angle. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 42 0.6691 9.8255 0.7431 9.8711 0.9004 9.9544 1.111 10.0456 48 10' .6713 .8269 .7412 .8699 .9057 .9570 1.104 .0430 50' 20' .6734 .8283 .7392 .8688 .9110 .9595 1.098 .0405 40' 30' .6756 .8297 .7373 .8676 .9163 .9621 1.091 .0379 30' 40' .6777 .8311 .7353 .8665 .9217 .9646 1.085 .0354 20' 50' .6799 .8324 .7333 .8653 .9271 .9671 1.079 .0329 10' 43 .6820 .8338 .7314 .8641 .9325 .9697 1.072 .0303 47 10' .6841 .8351 .7294 .8629 .9380 .9722 1.066 .0278 50' I 20' .6862 .8365 .7274 .8618 .9435 .9747 1.060 .0253 40' ft 1 30' .6884 .8378 .7254 .8606 .9490 .9772 1.054 .0228 30' " 1 40' .6905 .8391 .7234 .8594 .9545 .9798 1.048 .0202 20' 1 3 50' .6926 .8405 .7214 .8582 .9601 .9823 1.042 .0177 10'* 44 .6947 .8418 .7193 .8569 .9657 .9848 1.036 .0152 46 10' .6967 .8431 .7173 .8557 .9713 .9874 1.030 .0126 50' 20' .6988 .8444 .7153 .8545 .9770 .9899 1.024 .0101 40' 30' .7009 .8457 .7133 .8532 .9827 .9924 1.018 .0076 30' 40' .7030 .8469 .7112 .8520 .9884 .9949 1.012 .0051 20' 50' .7050 .8482 .7092 .8507 .9942 .9975 1.006 .0025 10' 45 .7071 .8495 .7071 .8495 1.0000 .0000 1.0JDO .0000 45 Nat. Log. Nat. Log. Nat. Log. Nat. Log. A _ _,!_ Cosine. Sine. Cotangent. Tangent. Angle. For angles over 45 use right column and take names of functions at bottom of page. TABLES 297 IV. NAPERIAN LOGARITHMS OF NUMBERS 1 to 9.9 No. 1 1 3 4 5 6 7 8 9 1 0.0000 0.0953 0.0182 0.2624 0.3365 0.4055 0.4700 0.5306 0.5878 0.6419 2 0.6931 0.7419 0.7885 0.8329 0.8755 0.9163 0.9555 0.9933 1.0296 1.0647 3 1.0986 1.1314 1 . 1632 1.1939 1.2238 1.2528 1.2809 1.3083 1.3350 1.3610 4 1.3863 1.4110 1.4351 1.4586 1.4816 1.5041 1.5261 1.5476 1.5686 1.5892 5 1.6094 1.6292 1.6487 1.6677 1.6864 1.7047 1.7228 1.7405 1.7579 1.7750 6 1.7918 1.8083 1.8245 1.8406 1.8563 1.8718 1.8871 1.9021 1.9169 1.9315 7 1.9459 1.9601 1.9741 1.9879 2.0015 2.0149 2.0281 2.0412 2.0541 2.0669 8 2.0794 2.0906 2.1041 2.1163 2.1282 2.1401 2.1518 2.1633 2.1748 2.1861 9 2.1972 2.2083 2.2192 2.2300 2.2407 2.2513 2.2618 2.2721 2.2824 2.2925 1 2.3026 2.3979 2.4849 2.5649 2.6391 2.7081 2.7726 2.8332 2.8904 2.9444 2 2.9957 3.0445 3.0910 3.1355 3.1781 3.2189 3.2581 3.2958 3.3322 3.3673 3 3.4012 3.4340 3.4657 3.4965 3.5264 3.5553 3.5835 3.6109 3.6376 3.6635 4 3.6889 3.7136 3.7377 3.7602 3.7842 3.8067 3.8286 3.8501 3.8712 3.8918 5 3.9120 3.9318 3.9512 3.9703 3.9890 4.0073 4.0254 4.0431 4.0604 4.0775 6 4.0943 4.1109 4.1271 4.1431 4.1589 4.1744 4.1897 4.2047 4.2195 4.2341 7 4.2485 4.2627 4.2767 4.2905 4.3041 4.3175 4.3307 4.3488 4.3567 4.3694 8 4.3820 4.3944 4.4067 4.4188 4.4308 4.4427 4.4543 4.4659 4.4773 4.4886 9 4.4998 4.5109 4.5218 4.5326 4.5433 4.5539 4.5643 4.5747 4.5850 4.5951 V. CONVERSION TABLES Radians to degrees. Degrees to radians. Grades to degrees. Mils to degrees. 0.1 5 44' 10' 0.00291 0.1 5.4' 1 3' 22. 5" 0.2 11 28 20 0.00582 0.2 10.8 2 6 45.0 0.3 17 11 30 0.00873 0.3 16.2 3 10 7.5 0.4 22 55 1 0.01745 0.4 21.6 4 13 30.0 0.5 28 39 2 0.03491 0.5 27.0 5 16 52.5 0.6 34 23 3 0.05236 0.6 32.4 6 20 15.0 0.7 40 06 4 0.06981 0.7 37.8 7 23 37.5 8 45 50 5 0.08727 0.8 43.2 8 27 00.0 0.9 51 34 10 0.17453 0.9 48.6 9 30 22.5 1.0 57 18 20 0.34907 1.0 54.0 10 33 45.0 2.0 114 35 30 0.52360 2.0 1 48.0 15 50 37.5 3.0 171 53 40 0.69813 3.0 2 42.0 20 1 7 30.0 4.0 229 11 50 0.87266 4.0 3 36.0 25 1 24 22.5 5.0 286 29 57 18 1.00000 5.0 4 30.0 30 1 41 15.0 6.0 343 46 60 1.04720 6.0 5 24.0 35 1 58 7.5 7.0 401 04 90 1.57080 7.0 6 18.0 40 2 15 00.0 8.0 458 22 8.0 7 12.0 50 2 48 45.0 9,0 515 40 9.0 8 06.0 60 3 22 30.0 10.0 9 00.0 70 3 56 15.0 20.0 18 00.0 80 4 30 00.0 30.0 27 00.0 90 5 3 45.0 40.0 36 00.0 50 45 00 100.0 90 00.0 298 TABLES VI. FUNCTIONS OF ANGLES IN GRADES Since 100 grades equals a quadrant or 90, functions of angles greater than 100 grades can be found from this table by use of formulas in Chapter VIII for finding functions of all angles in terms of functions of angles less than 90. Sine. Cosine. Tangent. Cotangent. Angle, grades. Nat. Log. Nat. Log. Nat. Log. Nat. Log. 0.0000 00 1.000 10.0000 0.0000 00 00 00 100 1 .0157 8.1961 0.9997 9.9999 .01571 8.1962 63.66 11.8058 99 2 .0314 .4971 .9995 .9998 .0314 .4973 31.82 .5027 98 3 .0471 .6731 .9989 .9995 .0472 .6736 21.20 .3264 97 4 .0628 .7979 .9979 .9991 .0629 .7988 15.90 .2013 96 5 .0786 .8946 .9970 .9987 .0787 .8960 12.71 .1040 95 6 .0941 .9736 .9955 .9981 .0945 .9756 10.58 .0244 94 7 .1097 9.0403 .9940 .9974 .1104 9.0430 9.057 10.9570 93 8 .1254 .0981 .9922 .9966 .1263 .1015 7.916 .8985 92 9 .1409 .1489 .9901 .9957 .1423 .1533 7.026 .8467 91 10 .1564 .1943 .9876 .9946 .1584 .1997 6.314 .8003 90 11 .1719 .2354 .9851 .9935 .1745 .2419 5.729 .7581 89 12 .1874 .2727 .9822 .9922 .1908 .2805 5.242 .7195 88 13 .2028 .3070 .9774 .9909 .2071 .3162 4.828 .6838 87 14 .2181 .3387 .9759 .9894 .2235 .3493 4.474 .6507 86 15 .2335 .3682 .9723 .9878 .2401 .3804 4.166 .6197 85 16 .2487 .3957 .9685 .9861 .2567 .4095 3.895 .5905 84 17 .2639 .4214 .9645 .9843 .2736 .4370 3.655 .5629 83 18 .2790 .4456 .9603 .9824 .2905 .4632 3.442 .5368 82 19 .2940 .4684 .9559 .9804 .3076 .4880 3.251 .5120 81 20 .3091 .4900 .9510 .9782 .3249 .5118 3.077 .4882 80 21 .3239 .5104 .9460 .9759 .3424 .5345 2.921 .4655 79 B 22 .3388 .5299 .9408 .9735 .3600 .5563 2.778 .4438 78 I 23 .3535 .5484 .9354 .9710 .3778 .5773 2.647 .4227 77 4 24 .3681 .5660 .9298 .9684 .3959 .5976 2.526 .4024 76 g 25 .3827 .5828 .9239 .9656 .4142 75172 2.414 .3828 75 1 26 .3972 .5990 .9177 .9627 .4327 .6362 2.311 .3638 74 - s 05 27 .4115 .6144 .9114 .9597 .4515 .6547 2.215 .3453 73 S 28 .4258 .6292 .9049 .9566 .4705 .6726 2.125 .3274 72 29 .4400 .6434 .8981 .9533 .4899 .6901 2.042 .3100 71 30 .4540 .6571 .8910 .9499 .5096 .7072 1.962 .2928 70 31 .4680 .6702 .8837 .9463 .5294 .7238 1.889 .2762 69 32 .4817 .6828 .8764 .9427 .5498 .7402 1.819 .2598 68 33 .4955 .6950 .8686 .9388 .5704 .7562 1.753 .2438 67 34 .5091 .7068 .8608 .9349 .5914 .7719 1.690 .2281 66 35 .5225 .7181 .8527 .9308 .6128 .7873 1.632 .2127 65 36 .5358 .7290 .8443 .9265 .6346 .8TJ25 1.576 .1975 64 37 .5490 .7396 .8358 .9221 .6569 .8175 1.522 .1825 63 38 .5621 .7498 .8272 .9176 .6797 .8323 1.472 .1678 62 39 .5750 .7597 .8181 .9128 .7028 .8468 1.423 .1532 61 40 .5878 .7692 .8091 .9080 .7266 .8613 1.376 .1387 60 41 .6005 .7785 .7997 .9029 .7508 .8755 1.332 .1245 59 42 .6129 .7874 .7901 .8977 .7757 .8897 1.289 .1103 58 43 .6253 .7961 .7804 .8923 .7950 .9037 1.248 .0963 57 44 .6374 .8044 .7706 .8868 .8274 .9177 1.209 .0824 56 45 .6494 .8125 .7605 .8811 .8541 .9315 1.171 .0685 55 46 .6613 .8204 .7501 .8751 .8817 .9453 1.134 .0547 54 47 .6730 .8280 .7396 .8690 .9099 .9590 1.099 .0410 53 48 .6845 .8354 .7290 .8627 .9391 .9727 1.065 .0273 52 49 .6960 .8426 .7181 .8562 .9692 .9864 1.032 .0137 51 50 .7071 .8495 .7071 .8495 1.0000 10.0000 1.000 10.0000 50 Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angle, Cosine. Sine. Cotangent. Tangent. grades. For angles over 50 grades read angles on right and take names of columns at bottom of page. TABLES 299 VII. NATURAL SINES AND COSINES ANGLES EXPRESSED IN MILS Mila. Degrees. Minutes. Sine. Cosine. Mils. 1.0000 1600 50 2 48.75 .0490 .9988 1550 100 5 37.50 .0980 .9952 1500 150 8 26.25 .1467 .9892 1450 200 11 15.00 .1951 .9808 1400 250 14 03.75 .2430 .9700 1350 300 16 52.50 .2903 .9569 1300 350 19 41.25 .3369 .9415 1250 400 22 30.00 .3827 .9239 1200 450 25 18.75 .4273 .9040 1150 500 28 07.50 .4714 .8819 1100 550 30 56.25 .5141 .8577 1050 600 33 45.00 .5556 .8315 1000 650 36 33.75 .5957 .8032 950 700 39 22.50 .6344 .7731 900 750 42 11.25 .6716 .7410 850 800 45 00.00 .7071 .7071 800 Mils. Degrees. Minutes. Sine. Cosine. Mils. *. INDEX (Numbers refer to pages) Abscissa, 37 Absolute value, 66, 122 Acceleration, definition of, 215 derivative of velocity, 215 Aggregation, signs of, 1 Amplitude, measurement of, 123 Angle, between two curves, 215 eccentric, 185, 186 measurement of, 104 mil, 106 of depression, 107 of elevation, 107 radian, 105 vectorial, 120 Arc, length of, 268 Arithmetic progression, 220 Axes, coordinate, 37 major and minor, 181 Binomial formula, 15 series, 234 Calculation, methods of, 20 by geometric method, 22 by logarithms, 25 by interpolation, 31, 39, 42, 43, 113 by slide rule, 30, 31 Centroid, 264 Cologarithm, definition of, 26 Complex number, addition of, 123 definition of, 63 division of, 124 graphic representation, 122 Complex number, modulus of, 122 multiplication of, 123 subtraction of, 123 Conic, definition of, 177 center of, 189 confocal, 188 diameter of, 183 graphs of, 164, 165, 166, 178, 179 Coordinates, definition of, 38 abscissa, 37 number pairs, 53, 69, 70 ordinate, 38 polar, 120 Cosine law, 95 example, 97 Critical value, 207 determination of, 207 Cube root, table of, 284 Curve, area under, 256 slope of, 202 angle between two, 204 Departure, definition of, 263 Depression, angle of, 107 Derivative, definition of, 194 slope of a curve, 202 to define motion, 215 Diameter of a curve, 183 conjugate, 184 Differential, definition of, 263 Directrix, 177 Eccentricity, e, 177 Elevation, angle of, 107 301 302 INDEX Ellipse, definition of, 179 eccentric angle of, 185 equation of, 180 major and minor axes of, 180 Empirical formulas, 57, 244, 245 Equations, equal roots, 217 equivalence of, 169 exponential, 28 general form, second degree, 187 quadratic, 13, 155 roots of, 8, 139, 140, 141 simultaneous, 9, 13, 169 solution of, 8 Equilibrant, 264 Equilibrium, of particle, 132, 133 of rigid body, 133 Factoring, Euclidean method, 5 formulas relating to, 2, 11 Force, components of, 130 moment of, 130, 133 Function, average value of, 259 continuous, 65 definition of, 65, 69, 70 derivative of, 194 even, 241 expansion of, in series, 232 hyperbolic, 168 increasing, 207 implicit and explicit, 163 integral, 139 irrational, 168 linear, 152 logarithmic, 237, 244 maximum value of, 205 minimum value of, 205 odd, 241 * rational fractional, 167 roots of, 139 theorems on, 140 trigonometric, 72 zero and infinity of, 141 Geometric progression, 222 Grade, definition of, table of, 298 Graphs, 36 construction of, 54 of functions, 53 of trigonometric functions, 108 representation by, 36, 205 Gravity, center ol, 264 Haversine, definition of, 73, 115 Highest Common Factor, 5 Hyperbola, definition of, 181 conjugate, 182 diameter of, 184 eccentric angle of, 186 equation of, 165, 182 Imaginary Quantity, (i), 11, 63 Increment, 66, 195 Index Laws, 2 negative exponents, 2 zero exponent, 2, 11 Inequalities, 14 Infinity, 65 Infinitesimal, 65 Integral, definite, 254 indefinite, 254 table of, 276 ntegration, 250 formulas of, 251 by parts, 267 by substitution, 269 Interpolation, methods of, 31 by means of graphs, 39, 42, 43 double, 31 simple, 280 special forms of, 113 Latitude, definition of, 114 Latus rectum, definition of, 178 Limit, definition of, 64 quadrant, 78 theorems on, 66, 193 INDEX 303 Lines, angle between, 160 distance between, 160 division of, 158 equations of, 157 general form, 152, 157 normal form, 157 one-point-slope form, 157 slope-intercept form, 157 slope of, 38, 153 two-point form, 158 theorem on, 152 Logarithm, change of base, 237 characteristic of, 27 definition of, 25 graph of function, 244 idea of, 24 logarithmic paper, 246 mantissa, 27 modulus, 237 Naperian base, 297 of trigonometric functions, 288 rules for calculation, 26 semi-logarithmic paper, 248 table of numbers, 286 Lowest Common Multiple, 4 Maximum value of function, 205, 206, 208, 209 Mil, definition of, 106 table of, 299 Minimum value of function, 205 Motion, 214 Number, 63 complex, 63, 123 imaginary, 63 rational and irrational, 62 reciprocal of, 23, 34 square roots of, 24, 284 scalar, 128 vector, 123 Ordinate, 37 Parabola, 164, 177 Parameter, definition of, 134 Particle, definition of, 133 equilibrium of, 133 Points, coordinates of, 38 distance between, 158 Proportion, definition of, 45 theorems on, 46 Quadrant Limits, 78 Radian, definition of, 45 Radicals, formulas relating to, 11 Ratio, definition, 45 Reciprocal of number, 23, 34 Remainder Theorem, 143 Resultant, 264 Scalar, 128 definition of, 125 product, 128 Sequence, 64 arithmetic, 219 geometric, 219 Series, arithmetic, 220 binomial, 234 expansion of function in, 232 exponential, 236 geometric, 222 harmonic, 225 harmonic mean, 226 power, 232 ratio test of, 230 sum of terms of, 220 with complex terms, 231 Slide Rule, 30 rules for, 31 Slope, definition of, 38 by derivative, 202 - of line, 38, 154 Sine Law, 95 Solid of revolution, definition of, 257 volume of, 257 304 INDEX Speed, 214 Square root, 24 table of, 284 Synthetic Division, 142 Tables, conversion, 297 explanation of, 280 functions of angles in grades, 298 integrals, 276 logarithms of numbers, 286 Naperian logarithms, 297 natural and logarithmic func- tions, 289 natural sines and cosines in mils, 299 powers and roots, 284 Tangent Law, 99 example, 99 Theorems, geometrical, 16 Transformations, linear, 174 of origin, 175 of variable, 269 Transformations, rotation of axes, 174 Trigonometric Functions, addition theorems, 85 complement relations, 80 conversion formulas, 98 definitions, 73 fundamental formulas, 75 graphs of, 108 inverse, 90 Variable, 63 continuous, 63 dependent and independent, 65 Variation, 46 Vector, addition of, 126 definition of, 123 polygon of, 127 product of, 128 radius, 120 Work, definition of, 133, 261 of variable force, 261 UNIVERSITY OF CALIFORNIA LIBRARY Los Angeles This book is DUE on the last date stamped below. REC'D COL. LIB. JUN161S6S. - i OCT151979 Form L9-Series 444 UC SOUTHERN REGIONAL UWARYFMIUTY A 000933186 9